THE  LIBRARY 

OF 

THE  UNIVERSITY 

OF  CALIFORNIA 

EDUC' 

PSYCH. 

.  LIBRARY 

GIFT  OF 


Emily  Turner 


Digitized  by  tine  Internet  Arciiive 

in  2007  with  funding  from 

IVIicrosoft  Corporation 


iittpV/www.archive.org/detaiis/eiementsofaigebrOOwentrich 


Elements  of  Algebra. 


BY 


G.  A.  WENTWORTH,  A.M, 

PROFESSOR   OF   MATHEMATICS   IN   PHILLIPS    EXETER  ACADEMY. 


BOSTON: 

PUBLISHED   BY   GINN   &   COMPANY. 

1886. 


WENTWORTH'S 
SERIES    OF   MATHEMATICS. 


Primary  School  Arithmetic. 

Grammar  School  Arithmetic. 

Practical  Arithmetic. 

Practical  Arithmetic  {Abridged  Edition). 

Exercises  in  Arithmetic. 

Shorter  Course   in  Algebra. 

Elements  of  Algebra. 

Complete  Algebra. 

University  Algebra. 

Exercises  in  Algebra. 

Plane  Geometry. 

Plane  and  Solid   Geometry. 

Exercises  in  Geometry. 

PI.  and  Sol.  Geom.  and   PI.  Trigonometry. 

Plane  Trigonometry  and  Tables. 

PI.  and  Sph.  Trig.,  Surveying,  and  Tables. 

Trigonometry,  Surveying,  and  Navigation. 

Log.  and  Trig.  Tables  {Seven). 

Log.  and  Trig.  Tables  {Complete  Edition). 


Special  Circular  and  Terms  on  application. 

-ouc- 

rS/CH. 

LIBRARY 

GIFT 

COPYRIGHT.   1881,   BY   G.   A.   WENTWOR-r  H. 
J.   S.   GUSHING    &    CO.,    PRINTERS,    BOSTON. 


PREFACE.         p^c/^, 

THE  single  aim  in  writing  this  volume  has  been  to  make  an 
Algebra  which  the  beginner  would  read  with  increasing  in- 
terest, intelligence,  and  power.  The  fact  has  been  kept  constantly 
in  mind  that,  to  accomplish  this  object,  "the  several  parts  must  be 
presented  so  distinctly  that  the  pupil  will  be  led  to  feel  that  he 
is  masterhiff  the  subject.  Originality  in  a  text-book  of  this  kind 
is  not  to  be  expected  or  desired,  and  any  claim  to  usefulness  must 
be  based  upon  the  method  of  treatment  and  upon  the  number 
and  character  of  the  examples.  About  four  thousand  examples 
have  been  selected,  arranged,  and  tested  in  the  recitation-room, 
and  any  found  too  difficult  have  bdfen  excluded  from  the  book. 
The  idea  has  been  to  furnish  a  great  number  of  examples  for  prac- 
tice, but  to  exclude  complicated  problems  that  consume  time  and 
energy  to  little  or  no  purpose. 

In  expressing  the  definitions,  particular  regard  has  been  paid  to 
brevity  and  perspicuity.  The  rules  have  been  deduced  from  pro- 
cesses immediately  preceding,  and  have  been  written,  not  to  be 
committed  to  memory,  but  to  furnish  aids  to  the  student  in  fram- 
ing for  himself  intelligent  statements  of  his  methods.  Each  principle 
has  been  fully  illustrated,  and  a  sufficient  number  of  problems  has 
been  given  to  fix  it  firmly  in  the  pupil's  mind  before  he  proceeds 
to  another.  Many  examples  have  been  worked  out,  in  order  to 
exhibit  the  best  methods  of  dealing  with  different  classes  of  prob- 
lems and  the  best  arrangement  of  the  work  ;  and  such  aid  has 
been  given  ia  the  statement  of  problems  as  experience  has  shown 

076 


IV  PREFACE. 

to  be  necessary  for  the  attainment  of  the  best  results.  General 
demonstrations  have  been  avoided  whenever  a  particular  illustra- 
tion would  serve  the  purpose,  and  the  application  of  the  principle 
to  similar  cases  was  obvious.  The  reason  for  this  course  is,  that 
the  pupil  mast  become  familiar  with  the  separate  steps  from  par- 
ticular examples,  before  he  is  able  to  follow  them  in  a  general 
demonstration,  and  to  understand  their  logical  connection. 

It  is  presumed  that  pupils  will  have  a  fair  acquaintance  with 
Arithmetic  before  beginning  the  study  of  Algebra;  and  that  suffi- 
cient time  will  be  afforded  to  learn  the  language  of  Algebra,  and 
to  settle  the  principles  on  which  the  ordinary  processes  of  Algebra 
are  conducted,  before  attacking  the  harder  parts  of  the  book.  "Make 
haste  slowly"  should  be  the  watchword  for  the  early  chapters. 

It  has  been  found  by  actual  trial  that  a  class  can  accomplish 
the  whole  work  of  this  Algebra  in  a  school  year,  with  one  reci- 
tation a  day ;  and  that  the  student  will  not  find  it  so  difficult  as 
to  discourage  him,  nor  yet  so  easy  as  to  deprive  him  of  the  re- 
wards of  patient  and  successful  labor.  At  least  one-fourth  of  the 
year  is  required  to  reach  the  Chapter  on  Fractions;  but,  if  the  first 
hundred  pages  are  thoroughly  mastered,  rapid  and  satisfactory 
progress  will  be  made  in  the  rest  of  the  book. 

Particular  attention  should  be  paid  to  the  chapter  on  Factoring; 
for  a  thorough  knowledge  of  this  subject  is  requisite  to  success  in 
common  algebraic  work. 

Attention  is  called  to  the  method  of  presenting  Choice  and  Chance. 
The  accomplished  mathematician  may  miss  the  elegance  of  the  gen- 
eral method  usually  adopted  in  Algebras ;  but  it  is  believed  that 
this  mode  of  treatment  will  furnish  to  the  average  student  the  only 
way  by  which  he  can  arrive  at  an  understanding  of  the  principles 
underlying  these  difficult  subjects.  In  the  preparation  of  these 
chapters  the  author  has  had  the  assistance  and  cooperation  of  G.  A. 
Hill,  A.M.,  of  Cambridge,  Mass.,  to  whom  he  gratefully  acknowl- 
edges his  obligation. 


PREFACE.  7 

The  chapter  on  the  General  Theory  of  Equations  has  been  con- 
tributed by  Professor  H.  A.  Howe  of  Denver  University,  Colorado. 

The  materials  for  this  Algebra  have  been-  obtained  from  English, 
German,  and  French  sources.  To  avoid  trespassing  upon  the  works 
of  recent  American  authors,  no  American  text-book  has  been  con- 
sulted. 

The  author  returns  his  sincere  thanks  for  assistance  to  Rev.  Dr. 
Thomas  Hill ;  to  Professors  Samuel  Hart  of  Hartford,  Ct. ;  C.  H. 
Judson  of  Greenville,  S.C. ;  0.  S.  Westcott  of  Racine,  Wis. ;  G.  B. 
Halsted  of  Princeton,  N.J. ;  M.  W.  Humphreys  of  Nashville,  Tenn. ; 
W.  LeConte  Stevens  of  New  York,  N.Y. ;  G.  W.  Bailey  of  New 
York,  N.Y. ;  Robert  A.  Benton  of  Concord,  N.H. ;  and  to  Dr.  D.  F. 
Wells  of  Exeter.  He  has  also  the  pleasure  of  expressing  his  obli- 
gations to  Messrs.  J.  S.  Gushing  and  F.  E.  Bartley,  to  whose  superior 
taste  and  judgment  the  typographical  excellence  of  this  book  is  due. 

There  will  be  two  editions  of  the  Algebra:  one  of  350  pages, 
designed  for  high  schools  and  academies,  will  contaih  an  ample 
amount  to  meet  the  requirements  for  admission  to  any  college ; 
the  other  will  consist  of  the  Elementary  part  and  about  150  pages 
more,  and  will  include  the  subjects  usually  taught  in  colleges. 

Answers  to  the  problems  are  bound  separately  in  paper  covers, 
and  will  be  furnished  free  to  pupils  when  teachers  apply  to  the 
publishers  for  them. 

Any  corrections  or  suggestions  relating  to  the  work  will  be  thank- 
fully received. 

G.  A.  TT-ENTWORTH. 
Pun.LiPs  Exeter  Academy, 
May,  1S81. 


COTsTTE^TS. 


CHAPTER  I.     Definitions  : 

Quantity  and  number,  1 ;  numbers,  2 ;  algebraic  numbers,  4  ; 
factors  and  powers,  7 ;  algebraic  symbols,  9 ;  algebraic  expres- 
sions, 10;  axioms,  11  ;  exercise  in  algebraic  notation,  13;  simple 
problems,  14. 

CHAPTER  II.     Addition  and  Subtraction  : 

Addition  of  algebraic  numbers,  16  ;  addition  of  monomials,  17; 
addition  of  polynomials,  19 ;  subtraction  of  algebraic  numbers, 
20;  subtraction  of  monomials,  21 ;  subtraction  of  polynomials,  22  : 
parentheses,  25 ;  simplifying  algebraic  expressions  by  removing 
parentheses,  26 ;  the  introduction  of  parentheses,  27. 

CHAPTER  III.     Multiplication  : 

Multiplication  of  algebraic  numbers,  28  ;  law  of  signs  in  mul- 
tiplication, 29;  multiplication  of  monomials  by  monomials,  30; 
the  product  of  two  or  more  powers  of  a  number,  30 ;  multiplica- 
tion of  polynomials  by  monomials,  31;  multiplication  of  polyno- 
mials by  polynomials,  32;  special  cases  of  multiplication  ;  square 
of  the  sum  of  two  numbers ;  the  square  of  the  difference  of  two 
numbers;  the  product  of  the  sum  and  difference  of  two  numbers, 
37;  square  of  a  trinomial,  39  ;  the  product  of  two  binomials  of  the 
form  x  +  a  and  x  +  b,  40 ;  the  powers  of  binomials  of  the  form 
a  ±6,  42. 

CHAPTER  IV.    Division  : 

Division  of  algebraic  numbers,  44 ;  division  of  monomials  by 
monomials,  46 ;  division  of  powers  of  a  number,  46  ;  division  of 
polynomials  by  monomials,  48  ;  division  of  polynomials  by  poly- 
nomials, 49 ;  use  of  parentheses  in  division,  53  ;  special  cases  of 
division  ;  the  difference  of  two  equal  odd  powers  of  two  numbers 
divisible  by  the  difference  of  the  numbers,  54  ;   the  sum  of  two 


ALGEBRA.  Vll 

equal  odd  powers  of  two  numbers  divisible  by  the  sum  of  the  num- 
bers, 55 ;  the  difference  of  two  equal  even  powers  of  two  numbers 
divisible  by  the  difference  and  by  the  sum  of  the  numbers,  56 ; 
the  sum  of  two  equal  even  powers  of  two  numbers  when  each 
exponent  is  the  product  of  an  odd  and  an  even  factor  divisible  by 
the  sum  of  the  powers  expressed  by  the  even  factor,  57 ;  general 
definitions  of  addition,  subtraction,  multiplication,  and  division,  58. 

CHAPTER  V.     Simple  Equations  : 

Definitions,  59  ;  transposition  of  terms,  60;  solution  of  simple 
equations,  60 ;  problems  in  simple  equations,  62. 

CHAPTER  VI.     Factors  : 

Case  in  which  all  the  terms  of  an  expression  have  a  common 
simple  factor,  68 ;  case  in  which  the  terms  of  an  expression  can  be 
so  arranged  as  to  show  a  common  compound  factor,  69  ;  resolution 
into  binomial  factors  of  trinomials  of  the  form  of  x^-h  {a  +  h)x  +  ah, 
70 ;  of  the  form  of  x'^  —  {a  +  h)x  +  ah,  71 ;  of  the  form  of  x^  +  {a—h)x 
—  ah,  72;  of  the  form  of  x^  —  {a  —  h)x  —  ah,  73;  of  the  form  of 
x^  -\-2ax  +  0^,  74;  of  the  form  of  x^  —  2ax-\-a^,  75;  resolution 
of  expressions  of  the  form  of  two  squares  with  the  negative  sign 
between  them,  76 ;  resolution  of  the  difference  of  two  equal  odd 
powers,  78  ;  resolution  of  the  sum  of  two  equal  odd  powers,  78 ; 
resolution  of  the  sum  of  two  equal  even  powers,  when  possible, 
79;  resolution  of  trinomials  of  the  form  of  x*  +  x^y'^  +y*,  80;  of 
the  form  of  2x^  +  5ax  +  2a^,  81 ;  resolution  of  polynomials  which 
are  perfect  powers,  82  ;  resolution  of  polynomials  composed  of  two 
trinomial  factors,  83  ;  resolution  of  polynomials  when  a  compound 
factor  of  the  first  three  terms  is  also  a  factor  of  the  remaining 
terms,  84. 

CHAPTER  VII.     Common  Factors  and  Multiples  : 

Highest  common  factor,  88 ;  method  of  finding  the  highest 
common  factor  by  inspection,  88 ;  method  of  finding  the  highest 
common  factor  by  division,  90;  principles  upon  which  this  method 
depends,  90;  this  method  of  use  only  to  determine  the  compound 
factor  of  the  highest  common  factor,  92 ;  modifications  of  this 
method  required,  93 ;  lowest  common  multiple,  98 ;  method  of 
finding  the  lowest  common  multiple  by  inspection,  98 ;  method 
,  of  finding  the  lowest  common  multiple  by  division,  100. 


CONTENTS. 


CHAPTER  VIII.     Fractions: 

Reduction  of  fractions  to  their  lowest  terms,  103;  reduction 
of  fractions  to  integral  or  mixed  expressions,  106 ;  reduction  of 
mixed  expressions  to  the  form  of  fractions,  107  ;  reduction  of  frac- 
tions to  the  lowest  common  denominator,  110;  addition  and  sub- 
traction of  fractions,  112;  multiplication  of  fractions,  120;  division 
of  fractions,  122;  complex  fractions,  124. 

CHAPTER  IX.     Fractional  Equations: 

Reduction  of  fractional  equations,  130 ;  reduction  of  literal 
equations,  134. 

CHAPTER  X.    Problems  Producing  Fractional  Equations,  137. 

CHAPTER  XI.  Simultaneous  Equations  of  the  First  Degree, 
151  ;  elimination  by  addition  or  subtraction,  152;  elimination  by 
substitution,  154  ;  elimination  by  comparison,  155  ;  literal  simul- 
taneous equations,  159. 

CHAPTER  XII.  Problems  Producing  Simultaneous  Equations, 
166. 

CHAPTER  XIII.     Involution  and  Evolution: 

Powers  of  simple  expressions,  181 ;  law  of  exponents,  181 ; 
powers  of  a  binomial  when  the  terms  of  the  binomial  have  coejQSi- 
cients  or  exponents,  182 ;  powers  of  polynomials  by  the  binomial 
method,  182;  roots  of  simple  expressions,  184;  imaginary  roots, 
184  ;  square  roots  of  compound  expressions,  186  ;  square  roots  of 
arithmetical  numbers,  188;  cube  roots  of  compound  expressions, 
190;  cube  roots  of  arithmetical  numbers,  193. 

CHAPTER  XIV.     Quadratic  Equations: 

Pure  quadratic  equations,  196  ;  affected  quadratic  equations, 
198 ;  literal  quadratic  equations,  203 ;  resolution  of  quadratic 
equations  by  inspection,  206;  number  of  roots  of  an  equation, 
208 ;  formation  of  equations  when  the  roots  are  known,  209 ; 
determination  of  the  character  of  the  roots  of  an  equation  by  in- 
spection, 200  ;  determination  of  the  maximum  or  minimum  value 


ALGEBRA.  IX 


of  a  quadratic  expression,  211 ;  higher  equations  whicli  can  be 
solved  by  completing  the  square,  212 ;  problems  involving  quad- 
ratics, 214. 

CHAPTER  XV.     Simultaneous  Quadratic  Equations  : 

Solution  when  the  value  of  one  of  the  unknown  quantities  can 
be  found  in  terms  of  the  other,  219;  when  each  of  the  equations 
is  homogeneous  and  of  the  second  degree,  222 ;  when  the  equa- 
tions are  symmetrical  with  respect  to  the  unknown  quantities,  223 ; 
problems  producing  simultaneous  quadratics,  226, 

CHAPTER  XVI.     Simple  Indeterminate  Equations  : 

The  values  of  the  unknown  quantities  dependent  upon  each 
other,  228 ;  method  of  solving  an  indeterminate  equation  in  posi- 
tive integers,  228. 

CHAPTER  XVII.     Inequalities: 

Fundamental  proposition,  234;  an  inequality  reversed  by 
changing  the  signs,  234. 

CHAPTER  XVIII.     Theory  of  Exponents: 

Fractional  and  negative  exponents,  23G ;  the  meaning  of  a 
fractional  exponent,  237 ;  the  meaning  of  a  negative  exponent, 
237 ;  laws  which  apply  to  positive  integral  exponents  apply  also 
to  fractional  and  negative  exponents,  238 ;  exercise  with  mono- 
mials having  fractional  and  negative  exponents,  239 ;  exercises 
with  polynomials  having  fractional  and  negative  exponents,  240 ; 
radical  expressions,  242;  reduction  of  surds  to  their  simplest 
forms,  243 ;  comparing  surds  of  the  same  order,  244 ;  comparing 
surds  of  different  orders,  245 ;  addition  and  subtraction  of  surds, 
247 ;  expansion  of  binomials  when  the  terms  are  radical  expres- 
sions, 249 ;  rationalization  of  the  denominator  of  a  radical  expres- 
sion, 249  ;  imaginary  expressions,  251 ;  square  root  of  a  binomial 
surd,  252 ;  equations  containing  radicals,  255 ;  solution  of  an 
equation  with  respect  to  an  expression,  257;  reciprocal  equations, 
258. 

CHAPTER  XIX.     Logarithms  : 

Common  system  of  logarithms,  261  ;  the  characteristic  of  a 
logarithm,  263;    the  mantissa  of  a  logarithm,  234;  logarithm  of 


CONTENTS. 


a  product,  264 ;  logarithm  of  a  power  and  of  a  root,  265 ;  loga- 
rithm of  a  quotient,  266  ;  a  table  of  four-place  logarithms,  270 ; 
general  proofs  of  the  laws  of  logarithms,  276  ;  solution  of  an  expo- 
nential equation  by  logarithms,  277. 

CHAPTER  XX.     Hatio,  Pkopoiition,  a]jd  Vaeiatiok  : 

Ratio,  278;  commensurable  and  incommensurable  ratios,  279; 
theorems  of  ratio,  281 ;  proportion,  284;  theorems  of  proportion, 
284;  variation,  292 ;  direct  variation,  293;  inverse  variation,  293. 

CHAPTER  XXI.     Series  : 

Infinite  series,  299  ;  finite  series,  299 ;  converging  series,  300  ; 
arithmetical  series,  301 ;  geometrical  series,  308  ;  limit  of  the  sum 
of  an  infinite  geometrical  series,  313  ;  harmonical  series,  314. 

CHAPTER  XXII.     Choice.    Binomial  Theorem: 

Fundamental  principle,  317;  arrangements  or  permutations. 
320  ;  number  of  arrangements  of  n  different  elements  taken  all  at 
a  time,  320  ;  number  of  arrangements  of  n  different  elements  taken 
r  at  a  time,  321 ;  number  of  arrangements  of  n  elements  of  which 
p  are  alike,  q  are  alike,  etc.,  324  ;  number  of  arrangements  of  n 
different  elements  when  repetitions  are  allowed,  325  ;  selectioas  or 
combinations,  326  ;  number  of  selections  of  r  elements' from  n  liif- 
ferent  elements,  326  ;  the  number  of  ways  in  which  p  elements 
can  be  selected  fromp+  r  different  elements  the  same  as  the  num- 
ber of  way3  in  which  r  elements  can  be  selected,  327  ;  value  of  r 
for  which  the  number  of  selections  of  n  different  elements  taken 
r  at  a  time  is  the  greatest,  330 ;  the  number  of  ways  of  selecting 
r  elements  from  n  different  elements  when  repetitions  are  allowed, 
336  ;  the  number  of  ways  in  which  a  selection  can  be  made  from 
?i  different  elements,  337 ;  the  number  of  ways  in  which  a  selec- 
tion can  be  made  from  p  +  q  +  r  elements  of  which  p  are  alike, 
q  are  alike,  r  are  alike,  338  ;  proof  of  the  binomial  theorem  when 
the  exponent  is  positive  and  integral,  342;  general  formula  for  the 
expansion  of  {a  +  x)n,  343;  general  term  of  the  expansion  of  a 
binomial,  344 ;  greatest  term  of  the  expansion  of  a  binomial,  345  ; 
proof  of  the  binomial  theorem  when  the  exponent  is  fractional, 
346  ;  when  negative,  348  ;  applications,  349. 


elemej^ts  of  algebea. 


CHAPTER   I. 

Quantity  and  Number. 

1.  Whatever  may  be  regarded  as  being  made  up  of 
parts  like  the  whole  is  called  a  Quantity. 

2.  To  measure  a  quantity  of  any  kind  is  to  find  how 
many  times  it  contains  another  known  quantity  of  the 
same  kind. 

3.  A  known  quantity  which  is  adopted  as  a  standard  for 
measuring  quantities  of  the  same  kind  is  called  a  Unit. 
Thus,  the  foot,  the  pound,  the  dollar,  the  day,  are  units 
lor  measuring  distance,  weight,  money,  time. 

4.  A  dumber  arises  from  the  repetitions  of  the  unit  of 
measure,  and  shows  hoiv  many  times  the  unit  is  contained 
in  the  quantity  measured. 

5.  When  a.  quantity  is  measured,  the  result  obtained  is 
expressed  by  prefixing  td  the  name  of  the  unit  the  number 
which  shows  how  many  times  the  unit  is  contained  in  the 
quantity  measured  ;  arid  the  two  combined  denote  a  quan- 
tity expressed/ in  units.  Thus,  7  feet,  8  pounds,  9  dollars, 
10  days,  are  quantities  expressed  in  their  respective  units. 


ALGEBRA. 


When  a  question  about  a  quantity  includes  the  unit,  the 
answer  is  a  number ;  when  it  does  not  include  the  unit, 
the  answer  is  a  quantity.  Thus,  if  a  man  who  has  fifteen 
bushels  of  wheat  be  asked  kow  many  bushels  of  wheat  he 
has,  the  answer  is  the  numbe)\  fifteen ;  if  he  be  asked  how 
much  wheat  he  has,  the  answer  is  the  quantity,  fifteen 
bushels. 

A  number  answers  the  question,  How  many  ?  a  quantity, 
the  question,  How  much  ? 


Numbers. 

6.  The  symbols  which  Arithmetic  employs  to  represent 
numbers  are  the  figures  0,  1,  2,  3,  4,  5,  6,  7,  8,  9.  The 
natural  series  of  numbers  begins  with  0 ;  each  succeeding 
number  is  obtained  by  adding  one  to  the  preceding  number, 
and  the  series  is  infinite. 

7.  Besides  figures,  the  chief  symbols  used  in  Arithmetic 
are : 

4-  (read,  plus),  the  sign  of  addition. 

—  (read,  minus),  the  sign  of  subtraction. 

X  (read,  multiplied  by),  the  sign  of  multiplication. 

~-  (read,  divided  by),  the  sign  of  division. 

=  (read,  is  equal  to),  the  sign  of  equality. 

Exercise.  — '  Read : 

7  +  12-19.  8  +  3-5-    20-15+1. 

9-    4=    5.  24  +  6=    10  X    3. 

6  X    4-24.  14-7  +  5-      6  x    2. 

48--   3  =  16.  9x5-180--   4. 

8.  Any  figure,  or  combination  of  figures,  as  7,  28,  346, 
has  one,  and  only  one,  value.     That  is,  figures  represent 


NUMBERS. 


particular  numbers.  But  numbers  possess  many  general 
properties,  which  are  true,  not  only  of  a  particular  number, 
but  of  all  numbers. 

Thus,  the  sum  of  12  and  8  is  20,  and  the  difference  be- 
tween 12  and  8  is  4.  Their  sum  added  to  their  difference 
is  24,  which  is  twice  the  greater  number.  Their  differ- 
ence taken  from  their  sum  is  16,  which  is  twice  the  smaller 
number. 

9.  As  this  is  true  of  any  two  numbers,  we  have  this  gen- 
eral property  :  The  sum  of  two  numbers  added  to  their  differ- 
ence is  tivice  the  greater  number ;  the  diffei'ence  of  two  numbers 
taken  from  their  sum  is  tiuice  the  smaller  number.     Or, 

1.  (greater  number  -f  smaller  number)  +  (greater  number 

—  smaller  number)  =  twice  greater  number. 

2.  (greater  number  +  smaller  number)  —  (greater  number 

—  smaller  number)  =  twice  smaller  number. 

But  these  statements  may  be  very  much  shortened ;  for, 
as  greater  number  and  smaller  number  may  mean  any  two 
numbers,  two  letters,  as  a  and  b,  may  be  used  to  represent 
them;  and  2a  may  represent  twice  the  greater,  and  2b 
twice  the  smaller.     Then  these  statements  become : 

1.  (a +  6)  + (a- 6)  =  2a. 

2.  (a-\-b)-{a-b)==2b. 

In  studying  the  general  properties  of  numbers,  letters 
may  represent  any  numerical  values  consistent  with  the 
conditions  of  the  problem. 

10.  It  is  also  convenient  to  use  letters  to  denote  numbers 
which  are  unknown,  and  which  are  to  be  found  from  certain 
given  relations  to  other  known  numbers. 


ALGEBRA. 


Thus,  the  solution  of  the  problem,  "  Find  two  numbers 
such  that,  when  the  greater  is  divided  by  the  less,  the  quo- 
tient is  4,  and  the  remainder  3 ;  and  when  the  sum  of  the 
two  numbers  is  increased  by  38,  and  the  result  divided  by 
the  greater  of  the  two  numbers,  the  quotient  is  2  and  the 
remainder  2,"  is  much  simplified  by  the  use  of  letters  to 
represent  the  unknown  numbers. 

11.  The  science  which  employs  letters  in  reasoning  about 
numbers,  either  to  discover  their  general  properties,  or  to 
find  the  value  of  an  unknown  number  from  its  relations 
to  known  numbers,  is  called  Algebra. 

Algebraic  Numbers.  . 

12.  There  are  quantities  which  stand  to  each  other  in 
such  opposite  relations  that,  when  we  combine  them,  they 
cancel  each  other  entirely  or  in  part.  Thus,  six  dollars 
gain  and  six  dollars  loss  just  cancel  each  other;  but  ten 
dollars  gain  and  six  dollars  loss  cancel  each  other  only  in 
part.  For  the  six  dollars  loss  will  cancel  six  dollars  of  the 
gain  and  will  leave  four  dollars. 

An  opposition  of  this  kind  exists  in  assets  and  dehts,  in 
income  and  outlay,  in  motion  forwards  and  hacTcwards,  in 
motion  to  the  right  and  to  the  left,  in  time  before  and  aften^ 
a  fixed  date,  in  the  degrees  above  and  below  zero  on  a 
thermometer. 

From  this  relation  of  quantities  a  question  often  arises 
Avhich  is  not  considered  in  Arithmetic ;  namely,  the  sub- 
tracting of  a  greater  number  from  a  smaller.  This  cannot 
be  done  in  Arithmetic,  for  the  real  nature  of  subtraction 
consists  in  counting  bachwards,  along  the  natural  series  of 
numbers.  If  we  wish  to  subtract  four  from  six,  we  start 
at  SIX  in  the  natural  series,  count  four  units  backwards,  and 


ALGEBRAIC    NUMBERS. 


arrive  at  two,  the  difference  souglit.  If  we  subtract  six 
from  six,  we  start  at  six  in  the  natural  series,  count  six 
units  backwards,  and  arrive  at  zero.  If  we  try  to  subtract 
nine  from  six,  we  cannot  do  it,  because,  when  we  have 
counted  backwards  as  far  as  zero,  the  natwral  series  of 
numbers  comes  to  an  end. 

'^  13.  In  order  to  subtract  a  greater  number  from  a  smaller 
it  is  necessary  to  assuvie  a  new  series  of  numbers,  beginning 
at  zero  and  extending  to  the  left  of  zero.  The  series  to  the 
left  of  zero  must  ascend  from  zero  by  the  repetitions  of  the 
unit,  precisely  like  the  natural  series  to  the  right  of  zero ; 
and  the  opposition  between  the  right-hand  series  and  the 
left-hand  series  must  be  clearly  marked.  This  opposition 
is  indicated  by  calling  every  number  in  the  right-hand 
series  a  positive  number,  and  prefixing  to  it,  when  written, 
the  sign  -f-  ;  ^^nd  by  calling  every  number  in  the  left-hand 
series  a  negative  number,  and  prefixing  to  it  the  sign  — . 
The  two  series  of  numbers  will  be  written  thus: 


-4,-3,-2,  -1,      0,   +1,  +2,  +3,  +A, 

I        I I I I \ I        I        I 

If,  now,  we  wish  to  subtract  9  from  6,  we  begin  at  6  in 
the  positive  series,  count  nine  units  in  the  negative  direction 
(to  the  left),  and  arrive  at  —  3  in  the  negative  series.  That 
is,  6- 9  =  -3. 

The  result  obtained  by  subtracting  a  greater  number 
from  a  less,  when  both  are  positive,  is  always  a  negative 
number. 

If  a  and  b  represent  any  two  numbers  of  the  positive 
series,  the  expression  a  —  b  will  denote  a  positive  number 
when  a  is  greater  than  b ;  will  be  equal  to  zero  when  a  is 
equal  to  b ;  will  denote  a  negative  number  when  a  is  less 
than  b. 

If  we  wish  to  add  9  to   —  6,  we  begin  at  —  6,  in  the 


V 


5  ALGEBRA. 

negative  series,  count  nine  units  in  the  positive  direciioit 
(to  the  right),  and  arrive  at  +  3,  in  the  positive  series. 

We  may  illustrate  the  use  of  positive  and  negative  num- 
bers as  follows : 

^   —5       0 8 20   .; 

h       A  0  ' 

Suppose  a  person  starting  at  A  walks  20  feet  to  the  right 
of  A,  and  then  returns  12  feet,  where  will  he  be  ?  Answer  : 
at  C,  a  point  8  feet  to  the  right  of  A.  That  is,  20  feet  — 
12  feet  =  8  feet ;  or,  20  -  12  =  8. 

Again,  suppose  he  walks  from  A  to  the  right  20  feet,  and 
then  returns  20  feet,  where  will  he  be  ?  Answer:  at  A,  the 
point  from  which  he  started.     That  is,  20  —  20  =  0. 

Again,  suppose  he  walks  from  A  to  the  right  20  feet,  and 
then  returns  25  feet,  where  will  he  now  be  ?  Answer :  at 
D,  a  point  5  feet  to  tl^e  left  of  A.  That  is,  if  we  consider 
distance  measured  in  feet  to  the  left  of  A  as  forming  a 
negative  series  of  numbers,  beginning  at  A,  20  —  25  =  —  5. 
Hence,  the  phrase,  5  feet  to  the  left  of  A,  is  now  expressed 
bj  the  negative  number  —  5. 

14.  Numbers  provided  with  the  sign  +  or  —  are  called 
algebraic  numbers.  They  dvQ  unknown  in  Arithmetic,  but 
play  a  very  important  part  in  Algebra.  In  contradistinc- 
tion, numbers  not  affected  by  the  signs  -f  or  —  are  termed 
absolute  ntimbers. 

15.  Every  algebraic  number,  as  -}-  4  or  —  4,  consists  of 
a  sign  +  or  —  and  the  absolute  value  of  the  number ;  in 
this  case  4.  The  sign  shows  whether  the  number  belongs 
to  the  positive  or  negative  series  of  numbers  ;  the  absolute 
value  shows  what  place  the  number  has  in  the  positive  or 
negative  series. 


FACTORS    AND    POWERS. 


16.  When  no  sign  stands  before  a  number,  the  sign  +  is 
always  understood  ;  thus,  4  means  the  same  as  +  4,  a  means 
the  same  as  +  «•     But  the  sign  —  is  never  omitted. 

17.  Two  numbers  which  have,  one  the  sign  +  and  the 
other  the  sign  — ,  are  said  to  have  unlike  signs. 

18.  Two  numbers  which  have  the  same  absolute  values, 
but  unlike  signs,  always  cancel  each  other  when  combined  ; 
thus,  +4-4  =  0,  +a-a  =  0. 

19.  The  use  of  the  signs  -|-  and  — ,  to  indicate  addition 
and  subtraction,  must  be  carefully  distinguished  from  their 
use  to  indicate  in  which  series,  the  positive  or  the  negative, 
a  given  number  belongs.  In  the  first  sense,  they  are  signs 
of  operations,  and  are  common  to  both  Arithmetic  and  Al- 
gebra. In  the  second  sense,  they  are  signs  of  opposition^ 
and  are  employed  in  Algebra  alone. 

Factors  and  Powers. 

20.  When  a  number  consists  of  the  product  of  two  or 
more  numbers,  each  of  these  numbers  is  called  a  factor  of 
the  product. 

When  these  numbers  are  denoted  by  letters,  the  sign  X 
is  omitted  •  thus,  instead  of  a  X  5,  we  write  ab  ;  instead 
oi  aXhx  c,  we  write  ahe. 

The  expression  abc  must  not  be  confounded  with  a-\-b-\-c\ 
the  first  is  a  product,  the  second  is  a  sum.  If  a  =  2,  ^  =  3,' 
c  =  4,  then 

a^>c  =  2x3x4  =  24; 

a  +  5  +  c  =  2-f  3  +  4=    9. 

21.  Factors  expressed  by  letters  are  called  literal  factors ; 
factors  expressed  by  figures  are  called  nnmerical  factors. 


8  ALGEBRA. 

22.  A  known  factor  of  a  product  which  is  prefixed  to  an- 
other factor,  to  show  how  many  times  that  factor  is  taken, 
is  called  a  coefficient.  Thus,  in  1c,  7  is  the  coefficient  of  c ; 
in  lax,  7  is  the  coefficient  of  ax,  or,  if  a  be  known,  7a  is 
the  coefficient  of  x.  When  no  numerical  coefficient  occurs 
in  a  product,  1  is  always  understood.  Thus,  ax  means  the 
same  as  lax. 

23.  A  product  consisting  of  two  or  more  equal  factors  is 
called  a  power  of  that  factor. 

/ 

24.  The  index  or  exponent  of  a  power  is  a  small  figure  or 
letter  placed  at  the  right  of  a  number,)  to-show  how  ma^iy 
times  the  number  is  taken  as  a  factor.  Thus,  o},  or  simply 
a,  denotes  that  a  is  taken  once  as  a  factor ;  d?  denotes  that 
a  is  taken  twice  as  a  factor  ;  a^  denotes  that  a  is  taken  three 
times  as  a  factor  ;  and  a"  denotes  that  a  is  taken  n  times  as 
a  factor.  These  are  read :  the  first  power  of  a  ;  the  second 
power  of  a  ;  the  third  power  of  a  ;  the  n\h  power  of  a. 

c^  is  written  instead  of  aaa.  \ 

a"^  is  written  instead  of  aaa,  etc.,  repeated  n  times. 
The  meaning  of  coefficient  and  exponent  must  be  care- 
fully distinguished.     Thus, 

4a  '=a-\-a-\-a-\-a', 

a^  =  aXaXaXa. 

If  a=:3,  4a  =3  +  3  +  3  +  3  =  12. 

a*  =  3  X  3  X  3  X  3  =  81. 

25.  The  second  power  of  a  number  is  generally  called 
the  square  of  that  number  ;  thus,  a^  is  called  the  square 
of  a,  because  if  a  denote  the  number  of  units  of  length  in 
the  side  of  a  square,  a-  denotes  the  number  of  units  of 
surface  in  the  square. 


ALGEBRAIC    SYMBOLS. 


The  third  power  of  a  number  is  generally  called  the  cube 
of  that  number ;  thus,  a^  is  called  the  cube  of  a,  because 
if  a  denote  the  number  of  units  of  length  in  the  edge  of  a 
cube,  o?  denotes  the  number  of  units  of  volume  in  the  cube. 


Algebkaig  Symbols. 

26.  Known  numbers  in  Algebra  are  denoted  by  figures 
and  by  the  first  letters  of  some  alphabet ;  as,  a,  b,  c,  etc.  ; 
«',  b' ,  c',  read  a  prime,  b  prime,  c  prime,  etc. ;  «!,  b^,  Ci,  read 
a  one,  b  one,  c  one. 

Unknown  numbers  are  generally  denoted  by  the  last  let- 
ters of  some  alphabet ;  as,  x,  y,  z,  x ,  y',  /,  etc. 

27.  The  symbols  of  operations  are  the  same  in  Algebra 
as  in  Arithmetic.  One  point  of  difference,  however,  must 
be  carefully  observed.  When  a  symbol  of  operation  is  omit- 
ted in  the  notation  of  Arithmetic,  it  is  always  the  sym^bol 
of  addition ;  but  when  a  symbol  of  operation  is  omitted  in 
the  notation  of  Algebra,  it  is  always  the  symbol  of  mul- 
tiplicaiion.     Thus,  456  means  400  -{-50  +  6,  but  4a6  means 

4  X  a  X  5  ;  4|  means  4  + 1,  but  4^  means  4  X  ^• 

0  b 

28.  The  symbols  of  relation  are  =,  >,  <,  which  stand 
for  the  words,  "  is  equal  to,"  "is  greater  than,"  and  "  is  lesa 
than,"  respectively. 

29.  The  symbols  of  aggregation  are  the  bar,  | ;  the  vin- 
^;^    ,    culum, ;   the  parenthesis,  (  )  ;   the  bracket,  [  ] ;   and 


X 


the  brace,  \  \ .     Thus,  each  of  the  expressions,    , 

(^  +  y),  [^  +  y],  S^  +  yJ,  signifies  that  a;-f  y  is  to  be  treated 
as  a  single  number. 


10  ALGEBRA. 

30.  TliG  symbols  of  continuatioii  are  dots, ,  or  dashes, 

^  and  are  read,  "and  so  on." 

31.  The  symbol  of  deduction  is  .'. ,  and  is  read,  ■'hence,*' 
or  "therefore." 

.^  Algebraic  Expressions. 

32.  An  algebraic  expression  is  any  number  written  in 
algebraic  symbols.  Thus,  8  c  is  the  algebraic  expression 
for  8  times  the  number  denoted  by  c. 

7(2^  —  Sab  is  the   algebraic    expression   for  7  times  the 
N    square  of  the  number  denoted  by  a,  diminished  by  3  times 
the  product  of  the  numbers  denoted  by  a  and  b. 

33.  A  term  is  an  algebraic  expression  the  parts  of  which 
are  not  separated  by  th^  sign  of  addition  or  subtraction. 
Thus,  Sab,  5xi/,  Sab-^bxy  are  terms. 

34.  A  monomial  or  simple  expression  is  an  expression 
which  contains  only  one  term. 

35.  A  polynomial  or  compound  expression  is  an  expression 
which  contains  two  or  more  terms.  A  binomial  is  a  poly- 
nomial of  two  terms.  A  trinomial  is  a  polynomial  of  three 
terms. 

36.  Like  terms  are  terms  which  have  the  same  letters, 
*v   and  the  corresponding  letters  affected  by  the  same  expo- 
nents.     Thus,  la^cx^  and  —ba^coi^  are  like  terms;    but 
la^c:^  and  —  ba(?x^  are  unlike  terms. 

37.  The  dimensions  of  a  term  are  its  literal  factors. 

38.  The  degree  of  a  term  is  equal  to  the  number  of  its 
dimensions,  and  is  found  by  taking  the  sum  of  the  expo- 
nents of  its  literal  factors.  Thus,  oa:y'is  of  the  second 
degree,  and  bx^y^  is  of  the  sixth  degree. 


AXIOMS.  11 


39.  A  polynomial  is  said  to  be  homogeneous  when  all  its 
terms  are  of  the  same  degree.  Thus,  1a^~5  a?y  +  xyz  is 
homogeneous,  for  each  term  is  of  the  third  degree. 

40.  A  polynomial  is  said  to  be  arranged  according  to  the 
powers  of  some  letter  when  the  exponents  of  that  letter 
either  descend  or  ascend  in  order  of  magnitude.  Thus,  Zaa? 
—  ^hdi?—^ax-\-'^h  is  arranged  according  to  the  descend- 
ing powers  of  x,  and  85  —  Gao:  —  45a;^-f-3 aj^  is  arranged 
according  to  the  ascending  powers  of  x. 

41.  The  numerical  value  o£  an  algebraic  expression  is  the 
number  obtained  by  giving  a  particular  value  to  each  letter, 
and  then  performing  the  operations  indicated. 

42.  Two  numbers  are  reciprocals  of  each  other  when  their 
product  is  equal  to  unity.     Thus,  a  and  -  are  reciprocals. 


Axioms. 

43.    1.    Things  which  are  equal  to  the  same  thing  are 
equal  to  each  other. 

2.  If  equal  numbers  be  added  to  equal  numbers,  the 
sums  will  be  equal. 

3.  If  equal  numbers  be  subtracted  from  equal  numbers, 
the  remainders  will  be  equal. 

4.  If  equal  numbers  be  multiplied  into  equal  numbers, 
the  products  will  be  equal. 

5.  If  equal  numbers  be  divided  by  equal  numbers,  the 
quotients  will  be  equal. 

6.  If  the  same  number  be  both  added  to  and  suLcracted 
from  another,  the  value  of  the  latter  will  not  be  altered. 

7.  If  a  number  be  both  multiplied  and  divided  by  an- 
other, the  value  of  the  former  will  not  be  altered. 


12  ALGEBRA. 


Exercise  I. 

If  a  =  l,  5  =  2,  €  =  ^,  d  =  4:,  e  =  5,/=0,find  the  nu- 
mericar values  of  the  following  expressions  : 

1.  9a  +  26  +  3c  — 2/  -cd       4.   — —  H . 

'"  0  a  e 

2.  U-^a-U  +  bcM       5.    le^hcd-^. 

'Zac 

^Z.    %ahc~hcd-\-^Gde-%f.     6.    ab(? -^-hcj} -  dec^^p/l 
7.    e*  +  6e2Z^2_^5^-4e^S-4£?5s.     ':    : 


a 

9. 

10. 

^^h""    l-))  lo         c'~d^              ) 

c^-ho'    ''^'     ~'    c^  +  cc^+.c^2" 

In  simplifying  compound  expressions,  each  term  must  be 
reduced  to  its  simplest  form  before  the  operations  of  addi- 
tion and  subtraction  are  performed. 

Simplify  the  following  expressions  : 

13.  100  +  80  --  4.  ^  "■-  ^         15.   25  -45  X  ^)-i^0  --  5.V      ^' 

14.  75-25x2.        •,    ^.  16.    24-5x4-- 10  +  3. 

17.    (24  -  5)  X  (4 --10 +  3). 

Find  the  numerical  value  of  the  following  expressions,  in 
which  a  =  2,  h  =  10,  x  =  3,  y  =  5  : 

18.   xi/  +  4:ax2.       ^  20.   Sx-[-77/~-7  +  aXy. 

p9.   xi/-lbb-~-b.  ^  21.    6^»-8y-=-2yx5~25. 


dl 


A  A 


ALGEBRAIC    NOTATION.  13 

22."  (65-8y)-f-2yx5  +  25.  "^6  / i^ 

23.  (6^>-8y)--(2yx5)  +  25.  ^Ol/^ 

24.  6^-(8?/--2y)x5-2^.  6      " 

25.  ^h~{h—y)~Zx-\-hxy-^Vda.      f 


''/: 


Algebraic  Notation. 

26.  Express  the  sum  of  a  and  h. 

27.  Express  the  double  of  x. 

28.  By  how  much  is  a  greater  than  5  ? 

29.  If  X  be  a  whole  number,  what  is  the  next  number  above 

it?     i-1-   , 

30.  Write  five  numbers  in  order  of  magnitude,  so  that  x 

shall  be  the  middle  number. 

31.  What  is  the  sum, of  x-{-x-{-x  -\- written  a  times ?  ^b'X 

32.  If  the  product  be  xy  and  the  multiplier  x,  what  is  the 

multiplicand  ?^  "^  ^'  ^  '  '^ 

33.  A  man  who  has  a  dollars  spends  h  dollars ;  how  many- 

dollars  has  he  left?      —  ^ 

34.  A  regiment  of  men  can  be  drawn  up  in  a  ranks  of  b  men 

each,  and  there  are  c  men  over ;  of  how  many  men 
does  the  regiment  consist?  (j 

35.  Write,  the  sum  of  x  and  y  divided  by  c  is  equal  to  the 

product  of  a,  h,  and  in,  diminished  by  six  times  c,  and 

increased  by  the  quotient  of  a  divided  by  the  sum  of 

X  and  y. 
33.    Write,  six  times  the  square  of  n,  divided  by  m  minus  a, 

increased   by  five   h   into  the   expression  c  plus  d^ 

minus  a. 
37.    Write,  four  times  the  fourth  power  of  a,  diminished  by 

five  times  the  square  of  a  into  the  square  of  b,  and 

increased  by  three  times  the  fourth  power  of  b. 


14  ALGEBRA. 


Exercise  II. 

That  the  beginner  may  see  how  Algebra  is  employed  in 
the  solution  of  problems,  the  following  simple  exercises  are 
introduced : 

1.    John  and  James  together  have  ?6.     James  has  twice 
as  much  as  John.     How  much  has  each  ? 

Let  X  denote  the  number  of  dollars  John  has. 

Then  2x  =  nurrher  of  dollars  James  has, 

and  X  -{-  2x  =  number  of  dollars  both  have. 

But  6  =  number  of  dollars  both  have  ; 

or  Sx:=G. 

and  x~'2. 

Therefore,  John  has  82,  and  James  has  $4. 

'  •  2.  A  stick  of  timber  40  feet  long  is  saw^ed  in  two,  so  that 
one  part  is  two-thirds  as  long  as  the  other.  Required 
the  length  of  each  part. 

Let  3  X  denote  the  number  of  feet  in  the  longer  part. 

Then  2x  =  number  of  feet  in  the  shorter  part, 

and     -  3  a;  +  2  a;  =  number  of  feet  in  both  together. 

But  40  =  number  of  feet  in  both  together ; 

.-.  3a;  +  2a;  =40, 
or  5a;  =  40, 

and  x=    8. 

Therefore,  the  longer  part,  or  Sx,  is  24  feet  long  ;  and 
the  shorter,  or  2  rr,  is  16  feet. 

Note.  The  unit  of  the  quantity  sought  is  always  given,  and  only 
the  iiTiinl)er  of  such  units  is  required.  Therefore,  x  must  never  be  put 
for  money,  length,  time,  weight,  etc.,  but  always  for  the  required 
numter  of  specified  units  of  money,  length,  time,  weight,  etc. 

The  beginner  should  give  particular  attention  to  this  caution. 


PROBLEMS.  15 


3.  The  greater  of  two  numbers  is  six  times  the  smaller 

and  their  sum  is  35.     Kequired  the  numbers. 

4.  Thomas  had  75  cents.     After  spending  a  part  of  Lis 

money,  he  found  he  had  twice  as  much  l^iz  dS  he 
had  spent.     How  much  had  he  spent? 

5.  A  tree  75  feet  high  was  broken,  so  that  the  part  broken 

off  was  four  times  the  length  of  the  part  left  standing. 
Eequired  the  length  of  each  part. 

Four  times  the  smaller  of  two  numbers  is  three  times 
the  greater,  nnd  their  sum  is  63.  Required  thci  num- 
bers. 

7.  A  farmer  sold  a  sheep,  a  cow,  and  a  horse,  for  $216. 

He  sold  the  cow  for  seven  times  as  much  as  the 
sheep,  and  the  horse  for  four  times  as  much  as  the 
cow.     How  much  did  he  get  for  each  ? 

8.  George  bought  some  apples,  pears,  and  oranges,  for  91 

cents.  He  paid  twice  as  much  for  the  pears  as  for 
the  apples,  and  twice  as  much  for  the  oranges  as  for 
the  pears.    How  much  money  did  he  spend  for  each  ? 

9.  A  man  bought  a  horse,  wagon,  and  harness,  for  $350. 

He  paid  for  the  horse  four  times  as  much  as  for  the 
harness,  and  for  the  wagon  one-half  as  much  as  for 
the  horse.     What  did  he  pay  for  each  f 

10.  Distribute  $3  among  Thomas,  Richard,  and  Henry,  so 

that  Thomas  and  Richard  shall  each  have  twice  as 

much  as  Henry,  "^vv^^?  I^Xof'J  ^d^j  ^    7-^  ,lo  0  C^/tXC 

11.  Three  men.  A,  B,  and  C,  pay  $1000  taxes.     B  pays  4 

times  as  much  as  A,  and  C  an  amount  equal  to  the 
sum  of  what  the  pther  twf  pay.  How  igjach  does 
each  pay  ?  ''\,  ,-:  .  a         '^  ZJ^.  ^^ 


CHAPTER  II. 

Addition  and  Subtraction. 

44.  An  algebraic  number  wbicli  is  to  be  added  or  sub- 
tracted is  often  inclosed  in  a  parenthesis,  in  order  that  the 
signs  +  and  —  which  are  used  to  distinguish  positive  and 
negative  numbers  may  not  be  confounded  with  the  -j-  and 
—  signs  that  denote  the  operations  of  addition  and  subtrac- 
tion. Thus,  4-  4  4-  (—  3)  expresses  the  sum,  and  +  4  —  (—  3) 
expresses  the  difference,  of  the  numbers  -}-  4  and  ~  3. 

45.  In  order  to  add  two  algebraic  numbers,  we  begin  at 
ilie  place  in  the  series  which  the  first  number  occupies,  and 
count,  in  the  direction  indicated  hy  the  sign  of  the  second 
number,  as  many  units  as'  are  equal  to  the  absolute  value 
of  the  second  number.  Thus,  the  sum  of  +  4  +  (-|-  3)  is 
found  by  counting  from  +4  three  units  in  the  positive 
direction,  and  is,  therefore,  -j-  7 ;  the  sum  of  +  4  +  (—  3)  is 
found  by  counting  from  +4  three  units  in  the  negative 
direction,  and  is,  therefore,  +1. 

In  like  manner,  the  sum  of  —  4  +  (+-3)  is  —  1,  and  the 
sum  of -4  + (-3)  is -7.     That  is, 

(1)  +4  +  (+3)-7;  (3)    -4  +  (+3)  =  -l; 

(2)  +4  +  (-3)-i;  (4)    _4,4-(-3)  =  -7. 

I.  Therefore,  to  add  two  numbers  with  like  signs,  find 
the  snm  of  their  absolute  values,  and  prefix  the  common  sign  ^ 
to  the  sum. 

II.  To  add  two  numl?ers  with  unlike  mgns,  find  the  differ- 
enoe  of  their  absolute  values,  and  prefix  the  sign  of  the  greater 
number  to  the  difference. 


ADDITION. 


n/ 


EXEECISE    III. 

1.  +  16  +  (-  11)=  ^  3.  +  68  +  (-  79)  =  -/  / 

2.  -15  +  (-25)  =  '^0  4.  -7  +  (+4)=    "/^' 

6.  +33  +  (+18)=  ^/ 

6.  +378 +  (+709)4- (-592)=    V    • 

7.  A, man  has  $5242  and  owes  $2758.     How  much  is  he 

worth  ? 

8.  The  First  Punic  War  began  B.C.  264,  and  lasted  23 

years.     When  did  it  end  ?       -.:  '-i  / 

9.  Augustus  Caesar  was  born  B.C.-63,  and  lived-f77  years. 

When  did  he  die  ?  '  ^'   '^-  '-   - 

10.  A  man  goes  65  steps  forwards,  then  37  steps  backwards, 
then  again  48  steps  forwards.  How  many  steps  did 
he  take  in  all  ?  How  many  steps  is  he  from  where 
he  started  ? 

Addition  of  Monomials. 

46.  If  a  and  h  denote  the  absolute  values  of  any  two 
numbers,  1,  2,  3,  4  (§  45)  become : 

(1)  +a  +  (+5)  =  a  +  Zi;  (3)    -  a  + (+5)  =-a  +  5  ; 

(2)  +a  +  (-5)  =  a-Z';  (4)    - a-\-{-h)  =  - a-h. 

Therefore,  to  add  two  terms,  write  them  one  after  the  other 
with  unchanged  signs. 

It  should  be  noticed  that  the  order  of  the  terms  is  im- 
material. Thus,  -\-a-~h=-^h-\-a.  Ifa  =  8  and  h  =  12, 
the  result  in  either  case  is  —  4. 

47.  3a  +  5a  +  2a  +  6a  +  a  =  17a. 
~2tf  — 3tf-c-4^-8<?  =  — 18c. 

Therefore,  to  add  several  like  terms  which  have  the  same 


18  ALGEBRA. 


sign,  add  the  coefficients,  prefix  the  common  sign,  and  annei 
the  common  symbols. 

48.  7a-6a  +  lla  +  a  — 5a  — 2a  =  19a— 13a  =  6«. 

—  3a  — 15a— 7a  +  14a -2a  =  14a  — 27a  =  — 13a. 

Therefore,  to  add  several  like  terms  whicli  have  not  all 
the  same  sign,  find  the  diffo^ence  between  the  sum_  of  the 
positive  coefficients  and  the  sum  of  the-  negative  coefficients, 
prefix  the  sign  of  the  greater  sum,  and  annex  the  common 


4^.  5a  — 25  + 3a  =  8a -25. 

—  3a;r  +  8?/ +  9a:r  —  4 1?  =  6a.r  +  8y  —  4<?. 

Therefore,  to  add  terms  which  are  not  all  like  terms, 
combine  the  like  terms,  and  write  down  the  other  terms,  each 
preceded  by  its  proper  sign. 

Exercise  IV. 
L   5a5  +  (-5a5)=  &    7a5  +  (-5a5)  = 

2.  d>mx-\-{-2mx)-=  7.    120^^^  + (- 95??^^)  = 

3.  -13mn^  +  (-7mn^)=-     a    -  33a5»  +  (  llaZ/^)  = 

4.  -'ox^-\-{j\-Sx-)=  9.    -75^y4-(+20:r?/)  = 

5.  25my2+(- 137723/2)=      lo.   +  15aV+(-a2:i-') -- 

11.  -h^m^+i-^lh'^m^)^ 

12.  5a  +  (-3Z>)  +  (+4a)  +  (-75)  = 

13.  4a2(?4-  (-  100,72)  +  (+  Q>a^c)  +  {-^X7jz) 

+  (-lla2.)  +  (+20:r7/.)  = 

14.  3a;2^  +  (-4a5)  +  (-277zn)  +  (+5.2.-2?/) 


ADDITION.  19 


Addition  of  Polynomials. 

50.  Two  or  more  polynomials  are  added  by  adding  their 
separate  terms. 

It  is  convenient  to  arrange  the  terms  in  columns,  so  that 
like  terms  shall  stand  in  the  same  column.     Thus, 

-3a3+    aH-ZaW-U^              ^o^y  +2 

2a3  +  2a2^»  +  6a62-3Z>8                 ;^y              ~    f 
2a«4-4a26  ^^^  -2c^y -5 


.  , ,  Exercise  V. 

Add: 

1.  5a  +  35  +  tf,     3a  +  3^>  +  3(^,     a  +  33-f-5c. 

2.  7a-45  +  e,     6a4-36-5tf,     -12a  +  4c. 
Z.   a-{-h  —  c,     b-\-c  —  a,     c-\~a  —  b,     a-\-b  —  c. 

4.  a  +  2Z>  +  3c,     2a-h-2c,     b-a~c,     c-a~b. 

5.  a-2^>  +  3c  — 4(f,     ^b-4:C-\-bd~2a, 

5tf-6t?+3a-4Z>,     7c?- 4a +  55 -4^. 

6.  5:8__4^Ji_j_5-p_3^     2a;3_Y^_7^_;^4^_^5^ 

7.  x''-2a^^2,x^,     a^j^^j^j;^     4a;*+5a^, 

2:i^  +  3a:-4,     -^x^^2x-b. 

8.  a8+3ai«-3a«5-Z/3,     2a3-j- 5^25  _  6a52_  7^2^ 

9.  2ab-^ax^-\-2a^x,     I2ab  ~^a^x-\-lOax\ 

aa^~Sab  —  ba^x. 


20  ALGEBRA. 


10.  e^-3c-3  +  2^-4c  +  7,     2c^-\-^(^-\-2c^-\-bc-\-e>, 

~4c^-4c2-5. 

11.  3^;^  — ;ry4"^2  — 3y^  +  4?/2  — 2^,    —ba^  —  xy  —  xz-\-byz, 

Q>a^-Q>y~Q>z,     4:ijz  -  b  yz -{- Z  z\ 

12.  w*  —  3  w^  w  —  6  ni^r?,     +  m^n^  -f  ?ri^7z^  —bvi^n, 

2m7i*  +  2n«  +  3m*,     -n*4-2w*+ 7m*w. 

Subtraction. 

51.  In  order  to  find  the  difference  between  two  algebraic 
numbers,  we  begin  at  the  place  in  the  series  luhich  the  Tninu- 
end  occupies,  and  count  in  the  direction  opposite  to  thai  indi- 
cated by  the  sign  of  the  subtrahend  as  many  units  as  are 
equal  to  the  absolute  value  of  the  subtrahend. 

Thus,  the  difference  between  +  4  and  +  3  is  found  by- 
counting  from  +4  three  units  in  the  negative  direction,  and 
is,  therefore,  + 1 ;  the  difference  between  +  4  and  —  3  is 
found  hj  counting  from  +  4  three  units  in  the  positive  direc- 
tion, and  is,  therefore,  +  7. 

In  like  manner,  the  difference  between  —  4  and  +  3  is 
—  7  ;  the  difference  between  —  4  and  —  3  is  —  1. 

Compare  these  results  with  results  obtained  in  addition : 


(1)  +4-(+3)  =  l 

(2)  +4-(-3)  =  7 

(3)  -4-(+3)  =  -7 

(4)  -4-(-3)  =  -l 

+  4  +  (-3)  =  l. 
+  4  +  (+3)  =  7. 
-4  +  (-3)  =  -7. 
-4  +  (+3)  =  -l. 

Or,     (1)   +4-(+3)  = 

(2)  +4-(-8)  = 

(3)  -4-(+3)  = 

(4)  -4-(-3)  = 

=  +  4  +  (-3). 
=  +  4  +  (+3). 
=  -4  +  (-3). 
=  -4  +  (+8). 

SUBTRACTION.  21 


52.  From  (1)  and  (3),  it  is  evident  tliat  subtracting  a 
positive  numher  is  equivalent  to  adding  an  equal  negative 
number. 

From  (2)  and  (4),  it  is  evident  that  subtracting  a  nega- 
tive number  is  equivalent  to  adding  an  equal  -positive  number. 

To  subtract,  therefore,  one  algebraic  number  from  another, 
change  the  sign  of  the  subtrahend,  and  then  add  the  subtra- 
hend to  the  Tninuend. 

\  Exercise  VI.  ^ 

1.  +25-(+16)=  3.   -    31-(+58)  = 

2.  -50 -(-25)=  4.   +  107 -(-93)- 

5.  Piome  was  ruled  by  emperors  from  B.C.  30  to  its  fall, 

A.D.  476.     How  long  did  the  empire  last? 

6.  The  continent  of  Europe  lies  between  36°  and  71°  north 

latitude,  and  between  12°  west  and  63°  east  longi- 
tude (from  Paris).  How  many  degrees  does  it  extend 
in  latitude,  and  how  many  in  longitude  ? 

»  Subtraction  of  Monomials. 

If  a  and  b  denote  the  absolute  values  of  any  two  num- 
bers, 1,  2,  3,  and  4  (§  51)  become : 

(1)  -{■a-{-\-b)^a~b.  (3)    -  a- {-\-b)  =  - a~  b. 

(2)  '\-a  —  {—b)==a-\-b.  (4)    ~  a  —  {—b)  =  — a-{-b. 

To  subtract,  therefore,  one  term  from  another,  change  the 
sign  of  the  term  to  be  subtracted  and  write  the  terms  one 
after  the  other. 


ALGEBRA. 


Exercise  \II. 

1.  dx-(-4:x)=  6.    l7aa^-(-24^a3^)-= 

2.  -^ab-(+5ab)=  7.    5a^x- (-3a^x)  = 

3.  SaP-(-\-10aP)=  8.    _4:ry- (-5a:y)  = 

4.  15m2:r2-(-7-w2a^)=      9.    8aa;-(-3ay)  = 

5.  -7mj-(-Sai/)=  10.    2aPy  -  {-\- ahy)  = 

11.  9.^-2  +  (5:^0 -(+ 8^')  = 

12.  5;r2y-(-18:rV)  +  (-10:z;2y)  = 

13.  17aa;3-(-a^)-(+24ar^)  = 

14.  —2>ah  +  (2mx)  —  {—4:mx)  = 

15.  3a-(+2Z^)-(-4c)  = 

Subtraction  of  Polynomials. 

53.  "Wlien  one  polynomial  is  to  be  subtracted  from  an- 
other, place  its  terms  under  the  like  terms  of  the  other, 
change  the  signs  of  the  subtrahend,  and  add. 

From  4^— SoT^y—    x\f^2if 

take  2;r^  —    ory  +  5  xy'^  —  3?/^ 

Change  the  signs  of  the  subtrahend  and  add  : 

^a^-Zx'y-    xif-\-2f 
-23?-{-    x'y-bxif  +  ^f 

2o(^-2c(^y-Q>xif-\-bf 

From  a^oi^-\-2a^x^  —  ^  ax* 

take  a^  +  4aV-3a^r^-4a5;* 


SUBTRACTION.  23 

In  the  last  example  we  have  conceived  the  signs  to  be 
changed  without  actually  changing  them.  The  beginner 
should  do  the  examples  by  both  methods  until  he  has  ac- 
quired sufficient  practice,  when  he  should  use  the  second 
method  only. 

Exercise  VIII. 

1.  YromGa-2b~ctsike2a~2b  —  Sc.     -' 

2.  YromSa-2b  +  Sctdke2a-1b-c  —  b, 

3.  From72^  —  8x  —  ltakeba^^—6x-]-3. 

4.  ¥rom4:X*-^a^-2x'-7x-{-9 

takex*-2a^-2a^-i-7x~0. 

5.  Yrom2x^-2ax  +  SaHsikea^-ax  +  a^. 

6.  From  c^ ~Zxy  —  y^ -\-yz  —  2^ 

take  3(?-\-2xy-\-hxz  —  Zf~2z^. 

7.  Froma»-3a25  +  3aZ>2-Z>3 

take  -  a«  +  3a25  -  3aZ-2  ^  h\ 

8.  YxQ>m3^  —  bxy-\-xz  —  y^-{-^yz-\-2^ 

take  x^  —  xy  —  xz-\-  2yz  +  3  2;^. 

9.  YxQm2a3?-^^abx-Wx-\-\2b^ 

take  aoi?  —  ^abx-\-bci?  —  ^Wx  —  :i:^. 

10.  From6a;3_Y^^^,_4^y_2^_5^_l_^y_4^_l_2 

\.^^8ci?-1o(?y^xif—if^9x^  —  xy-^G7f-~^. 

11.  Froma*-Z>*  take  ^a'^b -Ga^b^ -^^ab^,  and  from  the 

result  take  2a*  -^a?b-\-  Ga^b^  +  4a5^  —  25^ 

12.  From  .^^3/^  —  3  oc^'i^  +  4  :?;y*  —  3/^  take  —  ar*  +  2  :r*?/  —  4  a-?/* 

—  4^^.     Add  the  same  two  expressions,  and  subtract 
the  former  result  from  the  latter. 

13.  From  o?b'^  -  a^bc  -  8  ab'^c  -  a^c"  +  abe'-6  W^ 

take  2  a^bc  -  5  aWc  -f  2  a^c^  -  5  ^> V. 


24  ALGEBRA. 

14.  From    12a  +  3b  —  6c-2d   take    10a- Z> +  4c- 3c/, 

and  show  that  the  result  is  numerically  correct  when 
a  =  Q,  b  =  4:,  c=l,  d=b. 

15.  What  number  must  be  added  to  a  to  make  b  ;  and  what 

number  must  be  taken  from  2a?  —  Qa^b  -{-Q> ab^  —  2b^ 
to  \eB.Ye  a^- 7 a^b-Sb^? 

16.  From  2x^-y^-2xi/  +  z^  take  a^ -f-^2xi/ ~ z\ 

17.  From  12ac  +  8cd—9  take  —  7 ac  —  9c?c/ +  8. 

18.  From  —  Ga^  +  2a5  -  3/  take  4:a^ -{-6ab —  4:C^. 

19.  From  9a;y  — 4a:  — 3y+7  take  Sajy  —  2a;  + 3y  +  6. 

20.  From  —a^bc  —  ab^  c  +  abc^  —  abc 

take  a^5c?  +  «^^c  —  ohc^  +  a^c. 

21.  From  7^  — 2a; +  4  take  2^  + 3a;— 1. 

22.  From  ^a^  +  2a:y  —  if  take  —  a;^  —  3a;y  +  3y^,  and  from 

the  remainder  take  3a;^  +  4a;y  —  5y^. 

23.  From  aoi^  —  bf  take  cx^  —  dy^. 

24.  From  ax -\- bx  ^  by  -{-  cy  take  ax  —  bx  —  by-\-  cy. 

25.  From  5ar2  + 4a:  —  4y  +  3y2  take  bx^ —  ?>x^Zy -\-xf. 

26.  From  a^b^  +  \2abc  —  9ax^  take  ^ab^  —  6acx  +  Sa^a:. 

27.  From  a^ -2ab  +  e^ -SbHake  2a^-2ab  +  SP. 

28.  From  the  sum  of  the  first  four  of  the  following  expres- 

sions, a^  +  ^'  +  ^  +  cP,  d^-\-W-\-  c\  a"  ~  c"  -^b^-  d'\ 

^2  _  J2  _[_  ^2  _^  ^2^  ^2  _!_  ^  _|_  ^2  _  ^2^   ^^]^g  ^J^^  g^^^^  ^f  ^J^^ 

last  four. 

29.  From  2a^-2y^~z^  take  ^y^-\-2x^~z\  and  from  the 

remainder  take  32^  —  2y^  —  ar^. 

30.  From  a^  —  2a^c-\-^a(^  take  the  sum  of  a^ c?  —  2 a^  +  2 ac^ 

and  a^      ac?  —  d?c. 


parentheses.  25 

Parentheses. 
54.    From  (§  52),  it  appears  that 

(1)  a  +  {+h)=^a  +  h. 

(2)  a  +  (-b)  =  a-b. 

(3)  a~(-j-h)  =  a-b. 

(4)  a-(-^))  =  a  +  5. 

The  same  laws  respecting  the  removal  of  parentheses  hold 
true  whether  one  or  more  terms  are  inclosed.  Hence,  when 
an  expression  within  a  parenthesis  is  preceded  by  a  plus 
sign,  the  parenthesis  may  be  removed. 

When  an  expression  within  a  parenthesis  is  preceded  by 
a  minus  sign,  the  parenthesis  may  be  removed  if  the  sign 
of  every  tei'm  within  the  parenthesis  he  changed.     Thus  : 

(1)    a-\-{b  —  c)  =  a-\-h  —  c.\ 

55.  Expressions  may  occur  with  more  than  one  paren- 
thesis. These  parentheses  may  be  removed  in  succession, 
by  removing  yirsi!,  the  innermost  parenthesis ;  next,  the  inner- 
most of  all  that  remain,  and  so  on.     Thus : 

(1)    a-\b-{c-d)\ 
—  a~\b  —  c-{-dl, 
=  a  —  I^-}-c  —  d. 


(2)    a-[b-\e  +  (d-e-f)l] 
^a-[b-\c  +  (d-e+f)ll 
=  a-[b-le  +  d-e+f]l 
^=a  —  [b  —  c  —  d-{-e  — /], 
~a~  b-\-c-\-  d—  e  +/. 


26  ALGEBRA. 


Exercise  IX. 

Simplify  the  following  expressions  by  removing  the  paren- 
theses and  combining  like  terms. 

1.  (a  +  5)  +  (^  +  c)-(a  +  c). 

2.  (2a-b-c)-(a-2b-\-c). 

3.  (2x-7/)-(27/-z)-(2z-x). 

4.  (a  —  x  —  y)  —  (b  —  x-{-i/)  +  {c-\-2^). 

5.  (2x-i/  +  Sz)  +  (-x-7/-4:z)-(Sx-2,j-z). 

6.  (3a-  b  +  1  c)  -  (2a-i-3b)  -  (bb  -  4:c)  +  (Sc  -  a). 

7.  l-(l-a)-{-(l-a  +  a^)-(l-a  +  a^-a^). 

8.  a~l2b-(Sc  +  2b)-al. 

9.  2a-S5-(a-25)i. 

10.    Sa-\b-i-(2a-b)-(a-b)l 
11..  ■7a-[3a-S4a-(5a-2a)|]. 

12.  2a;  +  (y -  3^)  -K^x-  2y)  -\-z\-\-f>x-  (4y-  3z). 

13.  \{?>a-2b)-^(4:C-a)\-\a-{2b-Za)-c] 

-\-\a-{b-bc~a)\. 

14.  a-[2a  +  (3a-4a)]-5a-;6a-[(7a  +  8«.)-9a]|. 

15.  2a-{U-\-2c)-[bb-(fic-U)-\-bc 

-j2a-0:4-2i)|]. 

16.  6i-[2^»  +  ;3c-3a-(a  +  ^)|+!2a-(i  +  ^-)i]- 

17.  \^-x-[1x-\^x-{^x-3x-^x)\\ 


18.  2a  -  [35  +  (25 -c?)- 4^  + 12a -(35-^-25)!]. 

19.  a-[25  +  {3c-3a-(a  +  5)J  +  2a-(5-h3c)]. 

20.  a-[55-{a-(3c-35)  +  2(?-(a-25-^)J]. 


-/ 

/ 


PARENTHESES.  27 

56.  The  rules  for  introducing  parentheses  follow  directly 
from  the  rules  for  removing  them  : 

1.  Any  number  of  terms  of  an  expression  may  be  put 
within  a  parenthesis,  and  the  sign  plus  placed  before  the 
whole. 

2.  Any  number  of  terms  of  an  expression  may  be  put 
within  a  parenthesis,  and  the  sign  minus  placed  before  the 
whole ;  provided  the  sign  of  every  term  within  the  paren- 
thesis he  changed. 

It  is  usual  to  prefix  to  the  parenthesis  the  sign  of  the 
first  term  that  is  to  be  inclosed  within  it. 

Exercise  X. 

Express  in  binomials,  and  also  in  trinomials : 

1.  2a-35-4c4-c?H-3^-2/. 

2.  a  — 2a;  +  4y— 32:-2Z>-hc. 
8.    a'^  +  3a*-2a^-4a2  +  a-l. 

4.  -i^a-2h-\-2c-bd-e-~2f. 

5.  ax  —  hy  —  cz  —  hx -{- cy -{-  az. 

6.  2.i'^-3.z-V  +  4a;'y'-j5a;2^+^-22/^.  / 

7.  Express  each  of  the  above  in  trinomials,  having  the 

last  two  terms  inclosed  by  inner  parentheses. 

Collect-  in  parentheses  the  coefficients  of  x,  y,  z  in 

8.  2ax  —  &ay-\-4:hz  —  4:hx  —  '2cx  —  Zcy> 

9.  \ax-hx-\-2ay-\-Zy  +  ^az~Zbz~2z^    ■ 

10.  ^ax ~2by-^5cz  —  4:bx  —  Scy~{~ az  —  2cx~ay-\-4i hz.j 

11.  \2ax^\2ay-\-^hy'~\2lz  —  \bcx-\-^(yy-\-Zcz. 

12.  2ax  —  Zhy—  1  cz  ~  2bx  -\-  2cx -}-  8cz]-^2cx  ~  cy  ~  cz] 


CHAPTER  ni. 
Multiplication  of  Algebraic  Numbers. 

57.  The  operation  of  finding  the  sum  of  3  numbers,  each 
equal  to  5,  is  symbolized  by  the  expression,  3  X  5  =  15, 
read,  "  three  times  five  is  equal  to  fifteen  " ;  or,  by  the 
expression  5x3  =  15,  read,  "  five  multiplied  by  three  is 
equal  to  fifteen." 

58.  With  reference  to  this  operation,  this  sum  is  called 
the  product ;  one  of  the  equal  numbers  is  called  the  multi- 
plicand ;  and  the  number  which  shows  how  many  times  the 
multiplicand  is  to  be  taken  is  called  the  multiplier. 

59.  The  multiplier  means  so  many  times.  The  multipli- 
cand can  be  a  positive  or  a  negative  number  ;  but  the  mul- 
tiplier can  only  mean  that  the  multiplicand  is  taken  so 
many  times  to  he  added,  or  so  many  times  to  he  subtracted. 

60.  If  we  have  to  multiply  867  by  98,  we  may  put  the 
multiplier  in  the  form  100  —  2.  The  100  will  mean  that 
the  multiplicand  is  taken  100  times  to  be  added;  the  —  2 
will  mean  that  the  multiplicand  is  taken  twice  to  be  sub- 
tracted. 

In  general,  a  multiplier  with  +  before  it,  expressed  or 
understood,  means  that  the  multiplicand  is  taken  so  many 
times  to  be  added;  and  a  multiplier  with  —  before  it  means 
that  the  multiplicand  is  taken  so  many  times  to  be  sub- 
tracted.    Thus, 


MULTIPLICATION.  2^ 


(1)  +3x(+5)  =  (+5)  +  (+5)  +  (+5),or(+15). 

(2)  +3x(-5)  =  (-5)  +  (-5)  +  (-5),or(-15). 

(3)  _3x(+5)  =  -(+5)-(+5)-(+5),or(-15). 

(4)  -3x(-5)=-(-5)-(-5)-(-5),or(+15). 

From  these  four  cases  it   f<5llows,  that,  in  finding  the 
product  of  two  numbers, 

*61.    Like  signs  produce  plus  ;  unlike  signs,  minua. 


Exercise  XI. 

1.  - 

-17x8  = 

4. 

-18x-5  = 

2.   - 

-  12.8  X  25  = 

5. 

43  X- 

-6  = 

3.    - 

-3.29x5.49  = 

6. 

457  X 

100  = 

7. 

(-358-417)x- 

-79  = 

8.   (7.512  -  \~  2.8941)  X  (- 6.037  +  513.963 J)  = 

62.  The  product  of  more  than  two  factors,  each  preceded 
by  — ,  will  be  positive  or  negative,  according  as  the  number 
of  such  factors  is  even  or  odd.     Thus, 

-2x-3x-4=   +6x-4=   -24. 
-2x-3x-4x-5  =  -24x-5  =  +  120. 

a  13x8x-7  = 

10.  -38x9x-6  = 

11.  -20.9  X- 1.1x8  = 

12.  -  78.3  X  -  0.57  X  +  1.38  X  -  27.9  = 

13.  -  2.906  X  -  2.07G  x  -  1.49  X  0.89  =-- 


'60  ALGEBRA. 


Multiplication  of  Monomials. 

63.  The  product  of  numerical  factors  is  a  new  number 
in  whicli  no  trace  of  the  original  factors  is  found.  Thus, 
4  X  9  =  36.  But  the  product  of  literal  factors  can  only  be 
expressed  by  writing  them  one  after  the  other.  Thus,  the 
product  of  a  and  h  is  expressed  by  ab ;  the  product  of  ab 
and  cd  is  expressed  by  abed. 

64.  If  we  have  to  multiply  5  a  by  —  4  5,  the  factors  will 
give  the  same  result  in  whatever  order  they  are  taken. 
Thus,  5aX—4:b  =  bX—'^XaXb  =  —  20xab  =  —  20ab. 

65.  Hence,  to  find  the  product  of  monomials,  annex  the 
literal  factors  to  the  product  of  the  numerical  factors. 

66.  a^Xc^^=aaX  aaa  =  aaaaa  —  a*. 

a^  X  a^  X  a*  =:  aa  X  aaa  X  aaaa  =  aaaaaaaaa  =  a^. 

It  is  evident  that  the  exponent  of  the  product  is  equal  to 
the  sum  of  the  exponents  of  the  factors.     Hence, 

67.  The  product  of  two  or  more  powers  of  any  number  is 
ihat  number  with  an  exponent  equal  to  the  sum  of  the  expo- 
nents  of  th^  factors. 

Exercise  XII. 

1.  +aX  +  5  =  +  aZ>.  6.  —^pxSm  =  —  2^pm 

2.  -{-aX~b  =  —  ab.  7.  oa^X—a^  =  —  Sa^ 

3.  —aX-\-b  =  —  ab.  8.  —  3a  X  2a*  =  — 6a®. 

4.  —aX  —  b  =  -}-ab.  9.  6aX  — 2a  = 
6.  7aX^b  =  S5ab.  10.  577inX9m  = 


MULTIPLICATION. 


31 


11.  ZaxY.-^hy=  15.    5a'»X-2a'»-= 

12.  -8cmXc?w=  16.    Za^x'y.na^x^^ 

13.  -lahy^^ao^  17.    7aX-45x-8c  = 

14.  ^m^xY.Zm:f?=  la   Sa^^  X  Sac  X -4^2  = 

19.  27a&  X  — 39mp  X  18ap== 

20.  6a5y  X25y  X-5a2y  = 

21.  7m^;r  x3ma:^X  —  2mg'== 

22.  -?>pq'xQ>p^qX^p'(f  = 

23.  2  a^  m^o:'*  X  3  am^a.-^  X  4  a^ma;^  =      "^ 

24.  Qx'tj^X-^x'fz'x-2>x''yz  = 

25.  3aa:X  2amX  — 4ma;X62  = 

26.  7am2  y^  3 52^2 x  —  4a5  X  a^ Jn  X  —  2h^n  X  —  mn^  = 

Of  Polynomials  by  Monomials. 

68.   If  we  have  to  multiply  a-\-h  by  n,  that  is,  to  take 
{a-\-l))n  times  to  be  added,  we  have, 

{a-\-h)  Xn  —  {a-\-l))-\-(a-\-l)-\-{a-\-h)....n  times, 

=  a-\-a-\-a  .  .  .n  times  -\-h-\-h-\-h n  times, 

=  aXn'\-hXn, 
=  an-{-  hn. 

As  it  is  immaterial  in  what  order  the  factors  are  taken, 

7iX{a-\-  h)  =  an-\-  hn. 

In  like  manner, 

(^ci -{- h -{- c)  X  n  ==  an -\- hn -}- en, 
or,  n(a-{-h~{-c)  =  an-\-hn-\-  en. 


32  ALGEBRA. 


Hence,  to  multiply  a  polynomial  by  a  monomial, 

69.    Multiply  each  terin  of  the  polynomial  hy  the  monomial, 
and  add  the  partial  products. 


Exercise  XIII. 

1.  (6a  — 55)x3c  =  18ac-15Stf. 

2.  (2  +  3a-4a2_5a«)6a2==12a2-f  I8a^-24a*-30a* 

3.  ba{2>h-\-^c  —  d)  =  lbab-\-20ac  —  bad. 

4.  — 3aa7(—  hy  —  2cz-\-b)  =  Sahxy-\-6acxz  —  15ax. 

5.  (4a2-35)x3a5  = 

6.  {Sa^-9ab)xSa^=  --^ 

7.  (^a^-~4:f-}-5z^x2ar'y  = 

B.  (a^x-5a^x'  +  aa^  +  2x*)Xax'yr^ 

9.  (-9a*  +  3a352-4a253-5«)X-3a5*  = 

10.  (^a^-2ar^y-7xf  +  2/)x-5a^y=: 

11.  (-4a;3/*+5:r2^  +  82:«)X-3a;2^^ 

12.  (~S  +  2ab  +  a^b^X-a*  = 

13.  (-z-2:r/  +  5a:2^z2_g^yj^3^y^^)^_3^^2^ 

Of  Polynomials  by  Polynomials. 

70.    If  we  have  a -^  h -}- c  to  he  multiplied  hj  m-\-n  ~{-p, 
we  may  represent  the  multiplicand  a  +  5  +  c  by  M.     Then, 

M(m-{-n+p)^  MXm-j-  MXn  +  MXp. 

If  now  we  substitute  for  J/ its  value, 

(a  +  Z>  +  c)  {m  +  ?2  +p)  =  {a-\-h-\-c)Xm 
-\-la  +  b'\-c)Xn 
-\~(a  +  b-\~c)Xp; 


MULTIPLICATION.  33 


or,         (a-{~b-{-  c)  (in  +  n  -\-p)  =  aTn  +  hrn  +  cm 

-\-an  -\-hn  ■\- en 
+  ap  -\-h])  -\-cp. 

That  is,  to  find  the  product  of  two  polynomials, 

71.  Multiply  the  multiplicand  by  each  term  of  the  multiplier 
and  add  the  partial  products;  or,  m^ultiply  each  term,  of  one 
factor  by  each  term  of  the  other  and  add  the  partial  products. 

72.  In  multiplying  polynomials,  it  is  a  convenient  ar- 
rangement to  write  the  multiplier  under  the  multiplicand, 
and  place  like  terms  of  the  partial  products  in  columns. 

Thus:  (1)      5a  -    6   5 

3a  -   4   5 

Iba^-l^ab 

-20a5  +  24&^ 
15a2-38a^>  +  2452 

(2)  Multiply  4a;  +  3  +  5a:2_5^^y  4_(3^_5^_ 
Arrange  both  multiplicand  and  multiplier  according  to 

the  ascending  powers  of  x. 

3+    4:r+    bx"-    6:^8 

4-    bx-    ^x" 
\2-\-Ux-\-203^-2^a? 

-  15a;- 20a;2_  25^4_  30:r* 

-18a;^-24:g«-30a;^  +  36a^ 

12+      x-l^x^'-l^^a^  +36r^ 

(3)  Multiply  l  +  2:r  +  :?;^-3:^^by  ^^-2-2:1'. 
Arrange  according  to  the  descending  powers  of  x. 

x^-Zx^-\-2x  +1 

o^-1x  ~2 

a:^-3:c*  +  2:i-*+     3? 

-2a^  -\-Qx^-A:3^  —  2x 

-2x^  4-6:r2-4:r-2 


;^_5a:«  _|_7.^^+2:r2-6a:-2 


34  ALGEBRA. 


(4)    Multiply  a^-{-h^-\-(^  —  ab~hc  —  achja-\-h-{-c. 
Arrange  according  to  descending  powers  of  a. 

o?  —  ah  —  ac-\-    h^—      hc-{-    (? 

g  -f     ^+     ^ 

c?  —  o^h  —  (j?c-\-oh^~    abc-\-ac^ 

-^aH  -aV-    ahc  J^h^-h^c  +  hc^ 

-\-(^c —    ahc  —  a(?          -\- 1/ c  —  h(? -\- (^ 

The  student  should  observe  that,  with  a  view  to  bringing 
lik^terms  of  the  partial  products  in  columns,  the  terms  of 
the  multiplicand  and  multiplier  are  arranged  in  the  same 
order. 

In  order  to  test  the  accuracy  of  the  work,  interchange 
the  multiplicand  and  multiplier.  The  result  should  be  the 
same  in  both  operations. 

Exercise  XIV. 
Multiply : 

1.  a;2-4byrc2_{_5^  3^    «4_^^2^_|.^4  ^^^  ^2_^^ 

2.  y  —  6  by  y  +  13.  4.    o? -\- xy -{•  y^  hj  x  —  y . 

5.  2a:  — y  by  a: +  23/. 

6.  2a;3  +  4:?;2  +  8a:  +  16by  3.T-6. 

7.   x^-\'2i?-\-x  —  \\)Y  x—\.     8.    x^—Zax\>j  x-\-Za. 
9.    11r-\-Zah-a^hj -bh-\-na. 

10.  2a  +  Z>  bya  +  2Z».  12.    aV- a5  +  ^•^  by  a  +  ^>. 

11.  a^  +  aS  +  i^by  a-J.         13.    2a^»- SZ^^  by  3«2-4a6 

14.  -a«  +  2a2^>-53by  4a2  +  8a5. 

15.  c?-\-ab-\-l^hY  a^-ah-\-}?. 


MULTIPLICATION.  35 


16.  o?-Za'b  +  ?>ah''-¥hY  o'-^.ab-^-bK 

17.  x  +  2y  —  ZzhYX  —  2y  +  2>z. 

1&  2oi^  +  ^xy-\-4:'fhj^a^-4,xy-\-yz, 

19.  0!^ -\- xy -\- 'i/^  hj  o(^ -\- xz -\-  z^. 

20.  a^^h^-\-(^  —  ah  —  ac  —  hchY  a-\-h-\-c, 

21.  x^  —  xy  +  f^x  +  y  +  lhjx  +  y-l. 

Arrange  the  multiplicand  and  multiplier  according  to 
the  descending  powers  of  a  common  letter,  and  multiply  : 

22.  bx^^x'-\-x^-24:hY  x'+n-4:x. 

23.  ^+lla;-4a;2_24bya;2_|_5_|.4^^ 

24.  x'^-\-a^  —  4:X-ll-\-2c^hYa?-2x-{-Z. 

25.  -bx'^-x^-x-\-a^  +  l^x^hY  x^-2-2x. 

26.  ?>x-\-a^-2x^-^hY2x  +  ^x^-\-2,x'+l. 

27.  5a^  +  2a2^,2 _|_  ^js _  3^3^  ]t)y  Sa^S  -  2ah^-\-2>a^h^+h\ 

28.  4  aV  -  32  ay^  -  8  aV  + 1^  ^V  by  aV  +  4  «V*  +  ^  ^^V • 

29.  3m^+  3ri^  +  9mn^  +  9?/z^n  by  6?^^n^  —  2??2/i* 

—  6mW  +  2?^^w. 

GO.    Ga*^»  +  3a2Z.^  -  2aZ»«+  5«  by  4a^ -  2a^»3 _  3  J4^ 

Find  the  products  of : 

31.  a;  — 3,  x—1,  x-\-l,  and  x+2>. 

32.  :j.-2-a;+l,  ^■2-|-a;  +  l,  and  x^~x^+l. 

33.  a2_^a5  +  Z»^  a2_^^j_|_^2^  .^^^^  a^_aH^+h\ 

34.  4^3- 4^25 +  aZ^2^  4a24-3a^4-^,2^  and  2aH  +  h\ 

35.  2:  +  a,  .r  +  2 a,  a:  —  3 a,  a;  —  4a,  and  x-[-ba. 

36.  9a2  +  Z;2^  27a3_Z,3^27a«  +  ^',  and  Sla"" -^a'b^+b\ 


36  ALGEBRA. 


37.  From  the  product  of  2/^  —  2 yz—    z^  and  2/^-f-2yz—    2^ 

take  the  product  of  y^  —    yz  —  2z-  and  y^-\-    yz  —  2z^. 

38.  Find  the  dividend  when  the  divisor  =  3  a^  —  a5  —  3  6^, 

the  quotient  =a^5  — 26^,  the  remainder  =  — 2a6* 
-65^ 

The  multiplication  of  polynomials  may  be  indicated  by 
inclosing  each  in  a  parenthesis  and  writing  them  one  after 
the  other.  "When  the  operations  indicated  are  actually  per- 
formed, the  expression  is  said  to  be  smiplified. 

Simplify : 

39.  {a  +  h  —  c){a  -\-c~  b)(b  +  c-  a)(a  +  J  +  c). 

40.  (a  +  b)(b-\-c)-(c-{-  d)  {d-\-a)-{a  +  c)  {b  -  d). 

41.  (a-\-b^c-\-df-\-{a-b-c  +  dy 

+  la~b  +  c-df  +  (a-^b-c-  d)\ 

42.  {a-\-b-\-cf—a{b-^c~d)-b{a-{-c—b)~c{a-\-b—c). 

43.  {a-b)x-{b-c)a~\{b-x){b-a)-{b  —  c){b-\-c)\. 

44.  {m-\-n)rn  —  \{m  —  7if  —  {n  —  7ri)n\. 

45.  {a-b  +  cf-\a(c~a-b)-[b{a-\-b+c) 

—  c{a  —  b  —  cy\\. 

46.  {p^-\-(f)r-{p-\-q){p\r-q]-q\r-p]). 

47.  (9^:2^  _  4^4)  (^  _  y2)  _  J3:ry  -  2f\  \?>x{x^  +  y^) 

-2y(y2  +  3a:y-:r^)iy. 

48.  a^ ~-\2ab ~[-(a+\b -cWa-lb-cD-^^ab] 

-Uc\-(b  +  c)\ 

49.  5a(7-(a-5)(5  +  ^)S -^fS-(a-c)!. 

50.  5  \{a  —  b)x  —  cy\—  2\a{x  —  y)  —  bx\ 

-\Zax-{bc~2a)y\. 

51.  {x  -  l)(x  -  2)  -  3:r(;r  +  3)  +  2](x  +  2){x-\- 1)  -  3J . 


MULTIPLICATION.  3V 


52.    \{2a-^hf-{-{a-2bf\x\{d>a-2hy-{2a-Uf\. 
68.   4:{a-U){a  +  U)-2{a-^hf-2{o?-\-^b% 

54.  :^(^^J^ff-2a^f{x  +  y){x-y)-{:^-ff. 

55.  16  (a^  +  h^aj"  -  5^)  -  (2  a  -  3)(2  a  +  3)(4  a?  +  9) 

+  (25-3)(2^>  +  3)(4Z»2+9). 


73.  There  are  some  examples  in  multiplication  which 
occur  so  often  in  algebraical  operations  that  they  should 
be  carefully  noticed  and  remembered.  The  three  which 
follow  are  of  great  importance: 


(X)    a  -\-      b             (2)    a  -      b             (3)    a  +   b 

a  -\-      b                     a  —      b                     a  —   b 

a'+    ab                    a?-    ab                    a^^ab 

ab  +  b^                -    ab  +  b^                -oh- 

-b' 

a^^2ab-^b^             a^-2ab  +  b^             a? 

-b^ 

From  (1)  we  have  {a  -\-bf  =  a? -\-2ab-\-  b^.     That  is, 

74.  The  square  of  the  sum  of  two  numbers  is  equal  to  the 
sum,  of  their  squares  +  twice  their  product. 

From  (2)  we  have  (a  -by  =  a^-2ab -\-  W.     That  is, 

75.  The  square  of  the  difference  of  two  numbers  is  equal 
to  the  sum  of  their  squares  —  twice  their  product. 

From  (3)  we  have  {a -^b){a-h)  =  a^  —  b^.     That  is, 

76.  The  product  of  the  sum  and  difference  of  two  num- 
bers is  equal  to  the  difference  of  their  squares. 

11,    A  general  truth  expressed   by  symbols  is  called  a 
formula. 


38  ALGEBRA. 


78.  By  using  the  double  sign  i,  read  plus  or  minus,  we 
may  represent  (1)  and  (2)  by  a  single  formula ;  thus, 

in  which  expression  the  upper  signs  correspond  with  one 
another,  and  the  lower  with  one  another. 

By  remembering  these  formulas  the  square  of  any  bino- 
mial, or  the  product  of  the  sum  and  difference  of  any  two 
numbers,  may  be  written  by  inspection  ;  thus : 


Exercise  XV. 

1.  (127f-(123)«  =  (127-f  123)(127-  123) 

-=250x4  =  1000. 

2.  (29)2=  (30 -1)2=900 -60  + 1  =  841. 

3.  (53)2  =  (50 +  3)2  =  2500 +  300  + 9  =  2809. 

5.  {2a^x-bx'yf  =  ^a^x'-20o?:^y-\-2bx''f. 

6.  (3a62c  +  2a2c2)(3a62c-2a2c'2)  =  9a2  5V-4a*c' 


4^4 


7.  {x  +  yf=  15.  {ab  +  cdf  = 

&  (y_2)2=  16.  (3mn-4)2  = 

9.  (2a; +  1)2=  17.  (12  +  5a:)2  = 

10.  (2a  +  55)2=  18.  {^xf-y^f== 

11.  {l-3^f=  19.  (Sabc-bcdy  = 

12.  (Sax-4:x'y=  20.  {4:j^-xyy  = 

13.  (l-7a)2=  21.  {3:-i-y)(x-y)  = 

14.  (5;r?/  +  2)2=  22.  (2a^b)i2a-b)=^ 


MULTIPLICATION.  39 

23.  (S-x){S-\-x)  = 

24.  (Sah  +  2h^XSah-2b^)^ 

25.  (4:X^-Sf)(4:a^-i-S2/)  = 

26.  ((/r'-h7/^Xa^x'-{-by^y  = 

27.  (6xy  —  by^)(6xi/-\-bif)  = 

28.  (4r^-l)(4r^+l)  = 

29.  (l  +  Sah^Xl-dah^)  = 

30.  (aa;  +  ^3/)(«^^-%)(«^:r^+^V)  = 

79.    Also  the  square  of  a  trinomial  should  be  carefully 
noticed. 

a  -{-  h-j-  c 
a  -f-  h  -\-  c 
a*  +    ah-\-    ac 

ah  +^'+    ^tf 

ac  -\-    hc-\-  (^ 

a^  +  2ab-\-2ac  +  h^  +  2hc  +  (^, 
=-a'  +  h^-\-c'  +  2ah  +  2ac  +  2hc. 

It  is  evident  that  this  result  is  composed  of  two  sets  of 
numbers : 

I.    The  squares  of  a,  h,  and  c ; 
II.    Twice  the  products  of  a,  h,  and  c  taken  two  and  two. 

Again, 


a  — 
a  — 

h- 
h- 

c 
c 

a'- 

ah- 
ah 

ac 

-\-h^+    he 
ac         -\-    hc-\-(^ 

o?-2ab-2ac-\-h^-\-2hc^(^ 
^a^ -\-h^ ■\- c^ -2ah  -2ac -\-2 he. 


40  ALGEBRA. 


The  law  of  formation  is  the  same  as  before : 

I.    The  squares  of  a,  h,  and  c ; 
II,    Twice  the  products  of  a,  b,  and  c  taken  two  and  two. 

The  sign  of  each  double  product  is  +  or  —  according  as 
the  signs  of  the  factors  composing  it  are  like  or  unlike. 

The  same  law  holds  good  for  the  square  of  expressions 
containing  more  than  three  terms,  and  may  be  stated  thus : 

80.  To  the  sum  of  the  squares  of  the  several  terms  add  twice 
the  product  of  each  term  by  each  of  the  terms  that  folloxo  it. 

By  remembering  this  formula,  the  square  of  any  polyno- 
mial may  be  written  by  inspection  ;  thus : 

Exercise  XVI. 

1.  {x^y^zf-^  9.  (a3-f-^,3_|_^)2^ 

2.  {x-y-^zf=  10.  {:^-f-^f  = 

3.  {m-\-n-p  —  qf-=  11.  {x-^2y-Zzf-= 

4.  (a;2_|_2^_3)2=  12.  {^^^  _  ^f  ^  ^ ^)^  = 

5.  (f~^x^1)''=  13.  {y?^1x-2J=^    ' 

6.  (2x'-7x  +  9y=  14.  (x'-6x+7y  = 

7.  (x'-^-xf-z^y^  15.  (2x'-Sx-4:y=^ 

8.  (x^-Aa^y'  +  y^y^^  16.  (x-{-2y -\-3zy  = 

81.  Likewise,  the  product  of  two  binomials  of  the  form 
a;  +  a,  x-]-b  should  be  carefully  noticed  and  remembered. 


(1)     ^+5 

(2)     x-5 

X  +3 

X  -3 

a^  +  bx 

x^  —  i^x 

Sx+15 

-Sx+15 

ar^-f  8:r+15 

.r'-Sx  +  lb 

MULTIPLICATION. 

,^1 

(3)      X+6 

(4) 

X  -5 

V 

X  -3 

X  +S 

ar'-i-bx 

x'-bx 

-Sx-16 

+  3:r-15 

x'  +  2x-lb 

3^-2x-lb 

It  will  be  observed  that : 

I.  In  all  the  results  the  first  term  is  a^  and  the  last  term 
is  the  product  of  5  and  3. 

II.  From  (I)  and  (2),  when  the  second  terms  of  the  bino- 
mials have  like  signs,  the  product  has 

the  last  iexTR  positive ; 

the  coefficient  of  the  middle  term  =  the  snm  of  3  and  5  ; 
the  sign  of  the  middle  term  is  the  same  as  that  of  the 
3  and  5. 

III.  From  (3)  and  (4),  when  the  second  terms  of  the  bi- 
nomials have  unlike  signs,  the  product  has 

the  last  term  negative ; 

the  coeffilcient  of  the  middle  term  =  the  difference  of 
3  and  5; 

the  sign  of  the  middle  term  is  that  of  the  greater  of  the 
two  numbers. 

82.   These  results  may  be   deduced   from  the   general 

formula, 

{x -{- a){x -{-b)  =  x"  +  {a  +  b)  X  +  ah, 

by  supposing  for  (1)  a  and  b  bpth  positive ; 

(2)  a  and  b  both  negative ; 

(3)  a  positive,  b  negative,  and  a>b; 

(4)  a  negative,  b  positive,  and  a>b. 

By  remembering  this  formula  the  product  of  two  bino- 
mials may  be  written  by  inspection  ;  thus : 


42  ALGEBRA. 


Exercise  XVII. 

1. 

(x  +  2Xx  +  S)=                11. 

(.2;  —  (?)(:r  — c^)  = 

2. 

(x  +  l)ix  +  5)=                12. 

(x-AyXx  +  y)^^ 

3. 

(x-  3)(a;  --  6)  =                 13. 

(a-2b){a-5h)  = 

4. 

(x-8)(x~l)=                 14. 

{:^-\-2f){x^-\ff)  = 

5. 

(x-8Xx+l)=                 15. 

{f~Zxy){x^-\-xy)  = 

6. 

(x-2){xi-5)=                 16. 

{ax  -  ^){ax  +  6)  = 

7. 

(:^;-3)(:r+7)=  .                17. 

{x-\-a){x-h)^ 

8. 

(x-2)(x-4:)=                 18. 

(^-ll)(a;  +  4)=: 

9. 

(^+l)(:r+ll)=                 19. 

(^+12X^-11)  = 

10. 

(x-2a)(x  +  3a)-=           20. 

{x-W){x-b)  = 

83.    The  second,  third,  and  fourth  powers  oi  a-\-h  are 
found  in  the  following  manner : 


a  +       h 

a  +       b 

a^-\-    ab 

ah  +  y" 

(^:i^hj-- 

=  a^  +  2  ab  +  5^ 

a  +b 

■a^-{-2aH-^    aW 

aH-\-2ah'-  +  h^ 

{a-\-lSf-- 

=  a3  +  3a2^,-f-3a6-  +  Z^3 

a  +h 

a^J^ZaH  +  'daH^-\-    a¥ 

aH  +  ^aH^-\-?>a¥  +  b' 

{a-^by  =  a*  -{-  AaH  +  (jan^  -^  A  ab^  +  5* 


MULTIPLICATION.  43 


From  these  results  it  will  be  observed  that : 

I.  The  number  of  terms  is  greater  bj  one  than  the  ex- 
ponent of  the  power  to  which  the  binomial  is  raised. 

II.  In  the  first  term,  the  exponent  of  a  is  the  same  as 
the  exponent  of  the  power  to  which  the  binomial  is  raised ; 
and  it  decreases  by  one  in  each  succeeding  term. 

III.  b  appears  in  the  second  term  with  1  for  an  expo- 
nent, and  its  exponent  increases  by  one  in  each  succeeding 
term. 

IV.  The  coefficient  of  the  first  term  is  1. 

V.    The  coefficient  of  the  second  term  is  the  same  as 
the  exponent  of  the  power  to  which  the  binomial  is  raised. 

VI.  The  coefficient  of  each  succeeding  term  is  found 
from  the  next  preceding  term  by  multiplying  its  coefficient 
by  the  exponent  of  a,  and  dividing  the  product  by  a  num- 
ber greater  by  one  than  the  exponent  of  b. 

84.  If  b  be  negative,  the  terms  in  which  the  odd  powers 
of  b  occur  are  negative.     Thus  : 

(a  -  by  =  a'  -  4:a^b-{-6aH'  ~^a¥  +  b\ 


Exercise  XVIII. 
Write  by  inspection  the  results : 

1.  {x-\-af=              5.    (^  +  a)^=  '    9.  {x-\-yf^ 

2.  {x  —  o?)^=               6.    {x-af^  10.  {x-yj^ 

3.  {x  +  lf=               7.    {x-\-iy=  11.  (.r+l/  = 

4.  {x-\f=               8.    {x~iy^  12.  {x-lf=^ 


CHAPTER  IV. 
Division. 

85.  Division  is  the  operation  bj  whieli,  when  a  product 
and  one  of  its  factors  are  given,  the  other  factor  is  deter- 
mined. 

86.  "With  reference  to  this  operation  the  product  is  called 
the  dividend ;  the  given  factor  the  divisor ;  and  the  required 
factor  the  quotient. 

87.  The  operation  of  division  is  indicated  by  the  sign  -^; 

by  the  colon  : ,  or  by  writing  the  dividend  over  the  divisor 

12 
with  a  line  drawn  between  them.    Thus,  12  -^  4,  12  :  4,  — - , 

each  means  that  12  is  to  be  divided  by  4. 

88.  +  12  divided  by  -f  4  gives  the  quotient  -f-  3  ;  since 
only  a  positive  number,  +  3,  when  multiplied  by  -f-  4,  can 
give  the  positive  product,  +  12.  §  61. 

+  12  divided  by  --  4  gives  the  quotient  —  3  ;  since  only 
a  negative  number,  —  3,  when  multiplied  by  —  4,  can  give 
the  positive  product,  +  12.  §  61. 

—  12  divide^,  by  +  4  gives  the  quotient  —  3  ;  since  only 
a  negative  number,  —  3,  when  multiplied  by  +  4,  can  give 
the  negative  product  —  12.  §  61. 

—  12  divided  by  —  4  gives  the  quotient  -f-  3  ;  since  only 
a  positive  number,  +  3,  when  multiplied  by  —  4,  can  give 
the  negative  product,  -12.  §  61. 


DIVISION.  45 


(1)    ^  =  +  3.  (3)    ^  =  -3. 

+  4  +4 


(2)    ±^  =  -3.       '  (4)    -^J^'^^g 

—  4  —4 


From  (1)  and  (4)  it  follows  that 

89.  The   quotient   is  positive  when   the   dividend   and 
divisor  have  like  signs. 

From  (2)  and  (3)  it  follows  that 

The  quotient  is  negative  when  the  dividend  and  divisor 
have  unlike  signs. 

90.  The  absolute  value  of  the  quotient  is  equal  to  the 
quotient  of  the  absolute  values  of  the  dividend  and  divisor. 


Exercise  XIX. 


^    +264^  +3840^  106.33 

*     +4  •     -30  -4.9 


4648  ^     -2568  „    -42.435 

—  4.  — 


8  4-12  •  +34.5 


^'     +24  ^^'      +324    ~    ^ 


a  =:^=  11.       -1 


85  -  3.14159 


^    +6.8503^  ^     -.31831 


-61  -31.4159 


46 


ALGEBRA. 


Division  op  Monomials. 

91i  If  we  have  to  divide  ahc  by  he,  aahx  by  ahy,  12  abc 
by  —  4  a5,  we  write  them  as  follows : 

abc  aahx      ax  12  abc  o 

DC  aby        y  — 4a6 

Hence,  to  divide  one  monomial  by  another, 

92.  Write  the  dividend  over  the  divisor  with  a  line  between 
them;  if  the  expressions  have  common  factors,  remove  the 
common  factors. 

If  we  have  to  divide  a^  by  a^,  a^  by  a^,  a^  by  a,  we  write 
them  as  follows : 


a« 

aaaaa 

-aaa 

=0?, 

a' 

aa 

a« 

aaaaaa 

=  a\ 

a'  = 

aaaa 

=^aa 

a* 

aaaa 

«3 

— 

— 

aaa  - 

-  a . 

93.  That  is,  if  a  power  of  a  number  be  divided  by  a 
lower  power  of  the  same  number,  the  quotient  is  that  power 
of  the  number  whose  exponent  is  equal  to  the  exponent  of  the 
dividend  —  that  of  the  divisor. 

Again, 


a^         aa 

1         1 

a^  ~  aaaaa  ~ 

'  aaa~  0^^ 

a^        aaa 

1        1 

a^  ~~  aaaaa  ~ 

'  aa      a^ ' 

a*    aaaa 

1 

1 

a^  ~  aaaaaaaa  ~  aaaa  ~~ 

a* 

X 


DIVISION.  47 


94.  That  is,  if  any  power  of  a  number  be  divided  by  a 
higher  power  of  the  sanie  number,  the  quotient  is  expressed 
hy  1  divided  by  the  number  with  an  exponent  equxxl  to  the 
exponent  of  the  divisor  —  that  of  the  dividend. 

Exercise  XX. 
^    -\-ab       ,7  ^    \Oab  __  ,„   —^hmx 

1.    —  ~r  o.  /.    -— ; —  —  lo. —  = 

+  a  2bc  icu^ 

—  a  —a^  abc 

„   —ab_      7  g   —  12  am  _  m^p^x^ 

-\-a  —2m  mp^a? 


/. 

2bo 

8. 

x^ 

-cfi 

9. 

-12  am 

-2m 

Zbabcd 

LU. 

bbd 

11 

abx 

baby 

19 

21a' 

—  ab  _  I   1            JO    ^^^^^^  _  16    —  51  abdj/^  __ 

~a             '             '      bbd  '        2>bdy 

^  6w£_                  --      abx  _  22bm^y  __ 

2x                              baby  25  m?/^ 

-3a                           -3a3  -5a;3y 

iy»    "T — z ;; —  ^^* 


5  a*  nn^  x  —a^b^  cd^ 

^2x^1^%^  _  12am^nW^ 

7:ry2^  *     4:m^n^p^f 

23.    (4a2Jz3xl0a2  5«2)-5a3Z.V  = 

25.  104a5»:^»--(91a«5«a;^--7a^5*2:)  = 

26.  {2^an''x-~?>a^}y')  +  {^baH^x^^-baHx)-= 

27.  SSa*"*-^^-^  5a^"»*2  -  28.    84a*-^-i- 12a2  = 


48  ALGEBRA. 


Of  Polynomials  by  Monomials. 

95.    The  product  oi  (a  -{-  b  -{-  c)  X  p  =  ap  -}-  hp  -\-  cp. 
If  the  product  of  two  factors  be  divided  by  one  of  the 
factors,  the  quotient  is  the  other  faefeor.     Therefore, 

(op  -{-hp-j-  cpi)  ~-p  =  a-{-b-\-c. 

But  a,  5,  and  c  are  the  quotients  obtained  by  dividing 
each  term,  ap,  hp,  and  cp,  by  p. 

Therefore,  to  divide  a  polynomial  by  a  monomial, 


96.    Divide  each  terin  of  the  polynomial  by  the 


TnonoimaL 


Exercise  XXI. 

1.  {^ab~l2ac)-~^a  =  2b~?>c. 

2.  (15am  -  lO^m  +  20cm)  --  -  5m  =  -  3a  +  2 £  -  4c. 

3.  (18am?/-27^>ny  +  36cp3/)-^-9y  = 

4.  {2lax~l^bx+lbcx)-^-^x  = 

5.  (122^-8:r^  +  4a:)^4:r  = 

6.  (Z2(^-Q>a^  +  ^x'-\2x^)~Zx^^ 

7.  (35m^y  +  28m^y^— 14m?/^) -^ — 1my= 

8.  (4a*^-6a«^2_^12a2Z>3)^2a-6  = 

9.  {l2o(^2f-lbx^y^-24:X^y)~--?>x''y  = 

10.  {l2o^y^  -  24a;V'  +  ^^^f-  Ylx'f)  -^  Vlx'y'^ 

11.  {?>a''~2a^b-a^l?)~a^  = 

12.  (32;3y22  +  6a;V^  -  15a;^2/2  23+  18a;y  z)  -  -^Zx'yz  = 

13.  (-16a«Z»2^-^_^Sa^Z;26-''-12a^6^'2)---4a2Z>2c'2  = 


DIVISION.  49 


Of  Polynomials  by  Polynomials. 

97.    If  the  divisor  (one  factor)  =  a-\-b-}-  c, 

and  the  quotient  (other  factor)  =  71  -{-p  +  q, 

f      an  -\-bn-\-  en 
then  the  dividend  (product)     =\  -\- ap -\- bp -\- cp 

[-\-aq-\-bq  +  cq. 

The  first  term  of  the  dividend  is  an;  that  is,  the  product 
of  a,  the  first  term  of  the  divisor,  by  n,  the  first  term  of  the 
quotient.  The  first  term  n  of  the  quotient  is  therefore 
found  by  dividing  an,  the  first  term  of  the  dividend,  by  ft, 
the  first  term  of  the  divisor. 

If  the  partial  product  formed  by  multiplying  the  entire 
divisor  by  n  be  subtracted  from  the  dividend,  the  first  term 
of  the  remainder  ap  is  the  product  of  a,  the  first  term  of 
the  divisor,  by  p,  the  second  term  of  the  quotient.  That  is, 
the  second  term  of  the  quotient  is  obtained  by  dividing  the 
first  term  of  the  remainder  by  the  first  term  of  the  divisor. 
In  like  manner,  the  third  term  of  the  quotient  is  obtained 
by  dividing  the  first  term  of  the  new  remainder  by  the  first 
term  of  the  divisor,  and  so  on. 

Therefore,  to  divide  one  polynomial  by  another, 


98.  Divide  the  first  term  of  the  dividend  by  the  first  term 
of  the  divisor. 

Write  the  result  as  the  first  term  of  the  quotient. 

Multiply  all  the  terms  of  the  divisor  by  the  first  term  of 
the  quotient. 

Subtract  the  product  from  the  dividend. 

If  theix  be  a  remainder,  consider  it  as  a  new  dividend 
and  proceed  as  before. 


^^  ALGEBRA. 


99.  It  is  of  great  importance  to  arrange  both  dividend 
and  divisor  according  to  the  ascending  or  descending  pow- 
er's of  some  common  letter,  and  to  keep  this  order  throughout 
the  operation. 

Exercise  XXII. 
Divide 

(1)    c^  +  2ab  +  Phja-\-b-  (2)    a2-^,2|3ya  +  5; 
a'-^2ab-\-b^\a±h  a^ ~    P\a  +  b 

^!±__£^ a  +  b,  cU-ab_a~b 

abi-b^  -^ab-b^ 

ab  +  b^  -al-y^, 

(3)   a^-2ab-\-b''hj  a-b', 
a^~1ab^b^\a~b 


ah 


abJ^b^ 
ab-\-b^ 


(4)  4.a*x^~4:a^x*JrX^~a^hj  x'-a^- 
x^  -Aa'x'  +  A aV  -  a'^lx^  ~    a' 

^'-    a^x' x^-Sa^x'  +  a^ 

-Sa^x'  +  ia'x'-a^ 

-3aV  +  3a^^^ 

a*xr~  a^ 
a*x^  —  a^ 

(5)  22an'+15b'  +  Sa'-10a'b-22a¥hja'+Sb'-2ab; 
Sa*-10an  +  22a'b^-22ab^+15b'\   a'-2ab  +  Sb^ 

Sa*-    6a'b+    9a'b'' 3a-- 4aZ»-f- 5// 

—  4a^b  +  Ua^b^  -  22ab^ 
~   4:a^b+   8 a^ P- 12 ab^ 

ba'b^-lOab'+lbb* 
ba'b'-lOab^+l^b* 


DIVISION.  5J 


Divide 

6.   ar^—7x  +  12hjx  —  S. 

a   2x^-ar^  +  3x-9hj2x~S. 

9.   6a^+Ua^-4:x  +  24:hY2x  +  6. 
10.    3:i;2  +  a;  +  9^'^-l  by3.T-l. 
11>  7a;3_|_58^_24:r2-21  by  7a:-3. 

12.  :c^— 1  byo:— 1. 

13.  a3-2a^>2_j_^,3i3y^_5^ 

14.  x*  —  Sl7/hjx  —  Si/. 

15.  x^  —  y^  hj  X  —  7/. 

16.  a'' +  32^*  by  a +  25. 

17.  2a*  +  27a&3-81i^bya  +  35. 

1&    :r^  +  lla;2_i2:r-5:i^  +  6by3  +  A'2-3a7. 

19.  x*-9x^  +  a^-lQx-4:hy.z^  +  4:'^Ax. 

20.  36  +  ^*-132;2by6  +  2;2  +  52-. 

21.  x*  +  64by:r2  +  4:c  +  8. 

*^.  a;*  +  a,"3+57-35a:-24a;2by^.2_3  4.2a:. 

23.  l-2:-3:r2-:r'byl  +  2a;  +  a,-2. 

24.  a^-2:^+lhjx'-2x  +  l. 

25.  a*  +  2a2^-  +  9Z-nya2-2a5  +  352. 

26.  4ar'-:^+4«;by2  +  2x2^3^_ 

27.  a^- 243  by  a -3. 

28.  18:t'*  +  82:^2  +  40  -  67:^-45:2^  by  Ba^+6  -  4.r. 

29.  x*—6xi/  —  9a^  —  y^hyx^-f-y-rSx. 


^^  ALGEBRA. 


30.  x'-{-^:^y'~^x^y-^y^\,ja?~^xy-\.2y\ 

31.  x''^:^^J^y^\,j:^^^yj^yi 

32.  ^  +  ^  +  :rV  +  3/^-2.ry2_^yby:^  +  ,r-y. 

33.  '^^~^f-\-xy~xz~^yz~^hj2x^?>y-^z. 

34.  12  +  82.r-'  +  106a:^-70:r'-112:r«-38^ 

by  3- 5a: +7^^. 

35.  a;*  +  y'by:r*-:r«y  +  a;2^_^3^_|.y4_ 

36.  '^x''-^2x^f-2xf-n:^y~f},j2a?^f~xy. 

37.  16ar^  +  4a:2^+,/by4^_2^3/  +  3^^ 

38.  32a^5  +  8a3^«__«j5_4^2^4__5g^4^2 

by^»3_4^2^_^g^^^2_ 
^.    l  +  5;r«-6:rHyl-a;  +  3a;2. 

40.  \~b2a'h'~h\aWhY^aH^^2>ab~l. 

41.  ^V-^y'by:r33/  +  2:ryS-2:r2y2__^4_ 

42.  ^^+15:rV+15:r^2/^  +  3/«-6:r'y-6:ry'-20:r«y3 

by:r3-3:c2y  +  3a:?/2__y3 

43.  a'  +  2a3  3^-2a'»^^-2a«5-6a25^-3a£« 

by«3_2a2^-a52 

44.  81^y+18:r2y«-54r'/-18:r3y^-18^/-9/ 

by32;^  +  AY  +  y'. 

45.  «'  +  2a33  +  8a2^2_^  8aZ.«+165nya2  +  45'.      * 

46.  ^f~x^-V2\:^f-2^xifhjZxy~x^-f, 

47.  16aH9^H8a2^2|^y4a2_f.3^2_4^^_ 

48.  a^^h^J^c^-^cibchja-i-l^c. 

49.  «H8^3^c«_6a5cbyaH4iH^-ac-2a6~26c. 

50.  «^  +  ^'^  +  ^+3a2^  +  3a6Hya  +  5  +  c^. 


DIVISION.  53 


100.    The  operation  of  division  may  be  shortened  in  some 
cases  by  the  use  of  parentheses.     Thus : 

^  +  («  +  ^  +  ^) -^  +  (a5 -}- ac -{- he) X -{- abc \x  -\-h 


3^  +  (    +^       )^ 0!^  +  {a-{-c)x-\-ac 

{a        -{-c)3(^-\-{ab-\-ac-\-hc)x 
{a        -\-c)9^-{-{ab  -\-hc)x 

acx  +  <^^<? 

acx  -\-abG 


^   Exercise  XXIII. 
Divide 

1.  a^  (J  +  c)  +  h^  (a  -  c)  +  c^  (a  -  b)  +  abc  hj  a  +  h -\-  c. 

2.  a^  —  {a -\- b -\- c)  s? -\- {ab -\- ac -{-  be)  x  —  abc 

\>Y  x^  ~  {a -{- b)  X  ^  ab. 

^.01^-  2ax^-\-{a^-\-ab  -b^x-aH-\-  ab"^  by  :r  -  a  +  6. 

4.  x^  —  {c?  —  b  —  c)  y?  ~  {})  —  c)  ax  -\-bc  \)j  a?  —  ax  -\-  c. 

5.  'i/^—{7n-\-n -\-p)  y^  +  {mn -f  mp  +  np)y—  Tnnp by  y~p 

6.  ^H(5  +  a)^-(4-5a  +  ^)^-(4a  +  55):r  +  4  5 

^y  x^ -\- ^  X  —  4i. 

7.  x^-{a-^b  +  c-\-d)x^+{ab-\-ac+ad-\-bc-\-hd+cd)3^ 

—  (abc -\-  abd-\-  acd-{-bcd)  x-\-  abed 

by  a^~{a-\'c)x-\-ac. 

8.  ri*—  (m  —  c)x'^-\-{n  —  cm-\-d):^-\- (r  +  cn  —  dm) x^ 

+  (cr -\~  dn)  X -{- dr  hy  x^~msr^-i-nx-\-r. 

9.  x^  —  mx^  +  nx^  —  nx^  +  mo;  —  1  by  a:  —  1. 

10.    (x  +  yy  +  3(ix  +  yyz  +  3(x  +  y)z'  +  :^ 

^j(x  +  yy  +  2(x  +  y)z-{-z'. 


\ 
54  ALGEBRA. 


101.  There  are  some  cases  in  Division  which  occur  so 
often  in  algebraic  operations  that  they  should  be  carefully 
noticed  and  remembered. 


Case  I. 
The  student  may  easily  verify  the  following  results  : 

(1)   '^^  =  a?-{-ah-\-¥. 
a  —  b 


(2)   ^^''^~^^^  =  ^a'-\-^ah  +  U\ 
3a  —  2b 


(3)  ^L_=^  =  a'  +  a^b  +  a'  V  +  al?  +  ^>'. 

a  —  b 

(4)  'Lz^^  =  o>J^2a^b^^o?b^^%aW-^\U\ 

a  — 2b 


From  these  results  it  may  be  assumed  that : 

102.  The  difference  of  two  equal  odd  powers  of  any  two 
numbers  is  divisible  by  the  difference  of  the  numbers. 

It  will  also  be  seen  that : 

I.  The  number  of  terms  in  the  quotient  is  equal  to  the 
exponent  of  the  powers. 

II.  The  signs  of  the  quotient  are  all  positive. 

III.  The  first  term  of  the  quotient  is  obtained,  as  usual, 
by  dividing  the  first  term  of  the  dividend  by  the  first  term 
of  the  divisor. 

IV.  Each  succeeding  term  of  the  quotient  may  be  ob- 
tained by  dividing  the  preceding  term  of  the  quotient  by 
the  first  term  of  the  divisor,  and  multiplying  the  result  by 
the  second  term  'of  the  divisor  (disregarding  the  sign). 


DIVISION.  55 


Exercise  XXIV. 
Write  by  inspection  the  results  in  the  following  examples : 

1.  (2/3_i)_._(y_i).  5.    ^^J^_yS)^^^_yy 

2.  (^3- 125) -(6 -5).  6.    (a*-l)-{a-l). 

3.  (a^- 216) -(a -6).  7.   (1  -  Sa;^)  -  (1  -  2a:). 

4.  Cir^  -  343)  -  (.r  -  7).  a   {a^-?^2b')-^{x-2b). 

a    (8a^'c^-l)--(2«a;-l). 
10.    (l-27a;«2/^)- (1-3:^3/). 
U.    {^^a?If  -21  a^)  ^{^ab  -2>x). 

12.  (243a«-l)--(3a-l). 

13.  (32a* -2436^) --(2a- 3 i). 

Case  II. 

(1)   ^±^  =  a'-ah  +  b\ 
a  +  o 

27^!+M  =  9.r^-6:ry  +  43/». 

(3)  ^^-i^  =  a*-a^6  +  a2i^-a^'  +  6*. 

a  +  5 

(4)  243^  +  32,^*^.3-^^.4 _ 54^^^ ^ 36:r2^_ 24^^^_i- 167/*. 

3a;  +  2y 

From  these  results  it  may  be  assumed  that : 

103.  The  sum  of  two  equal  odd  powers  of  two  nuinbers  is 
divisible  by  the  sum  of  the  nuinbers. 

The  quotient  may  be  found  as  in  Case  I.,  but  the  signs 
are  alternately  plus  and  minus. 


56  ALGEBRA. 

Exercise  XXV. 
Write  by  inspection  the  results  in  the  following  examples  • 

1.  {a^-^^f)^{x-{-y)..  5.    {^a^:^ -\-l)-~  {2ax+l). 

2.  {:^-\-^)^{x  +  y).  6.   {:^ +  21  ^)-^  {x-\-2.y). 

3.  (l  +  8a«)-f-(l  +  2a).       7.   {c^ +  2>2¥)~{a  +  2h). 

4.  {21a^-]-W)~{Za  +  h).     8.    (512:r32/^  +  ^)-^(8:ry  +  2). 

9.  (729a«  +  216Z>3)--(9a  +  65). 

10.  (64a8+1000Z*3)-f-(4a  +  105). 

11.  (64  a^  h^  +  27.r3)  h-  (4  a5  +  3  a,-). 

12.  (a;3_^343)^(^_^7) 

13.  {21o(^f  4-  8  2^)  -^  (3 ^3/  4_  2 z). 

14.  (1024a* +  2435')- (4a +  3^>). 

Case  III. 

(1)    ^~y'  ^x  +  y.  (2)    ^— ^'  =  a,-^  +  a;2    +^^_l_^^^ 

x-y  x-y 

(3)    ^^'  =  :r-y.  (4)    '''-^^  :^  -  x'y  +  xf  -  y'. 

x  +  y  x-\-y 

From  these  results  it  may  be  assumed  that : 

104.  The  diffei^ence  of  two  equal  even  powers  of  two  num- 
bers is  divisible  by  the  difference  and  also  by  the  sum  of  the 
numbers. 

When  the  divisor  is  the  difference  of  the  numbers,  the 
quotient  is  found  as  in  Case  I. 

When  the  divisor  is  the  sum  of  the  numbers,  the  quotient 
is  found  as  in  Case  II. 


DIVISION.  57 


Exercise  XXVI. 
Write  by  inspection  the  results  in  the  following  examples : 

1.  {x'-y^)^{x-y).  a    {l^x^-l)~{2x-\-l). 

2.  {x''-y')~{x-\-y).  9.    {^la^x^-l)^{^ax-l). 

3.  {a^-x^)-r-{a-x).  10.    {Sla'x^-V)^{d>ax+l). 

4.  {a^-x^)^{a  +  x).  11.    {Q^a"" -W) -~{2a-h). 

5.  {x*-^U/)-~{x-?>y).       12.    (64a«-^«)--(2a  +  5). 

6.  {x""  -  81  y^)  -f-  (a:  +  3y).       13.    {x^  -  729/)  -  (a;  -  Sy). 

7.  (16:r^-l)--(2:r-l).        14.    (a;«  -  729/) -- (a;  +  3y). 

15.  (81a''-16c''')--(3a-26'). 

16.  (81a*- 16 c--*)- (3 a +  26). 

17.  (256  a*  ~  10,000)  --(4  a-  10). 

18.  (256  a*  -  10,000)  h-  (4  a  +  10). 

19.  (625  a,-* -1)- (5  a: -1). 

Case  IV. 

It  may  be  easily  verified  that : 

105.  The  sum  of  two  equal  even  poivers  of  iiuo  numheis 
is  not  divisible  by  either  the  sum  or  the  difference  of  the 
numbers. 

But  when  the  exponent  of  each  of  the  two  equal  powers 
is  composed  of  an  odd  and  an  even  factor,  the  sum  of  the 
given  powers  is  divisible  by  the  sum  of  the  powers  expressed 
by  the  even  factor. 

Thus,  x^  +  /  is  not  divisible  by  a;  -f-  y  or  by  a;  —  y,  but 
is  divisible  by  o?  +  y^. 

The  quotient  may  be  found  as  in  Case  II. 


58  ALGEBRA. 


Exercise  XXVII. 
Write  by  inspection  the  results  in  the  following  examples  : 

1.  {x'-{-f)-^{:^  +  i/).         6.    (^^'+l)-(:.^+l). 

2.  (a«+l)--(a2  +  l).  7.    (64.T«  +  y<') -- (4ar'  +  /). 

3.  (a^«  +  2/i°)--(a2  +  y2j        8,    (64 +  ««)-- (4  +  a^). 

4.  (^1^+1)-- (52+1).  9.    (729a«  +  5'')^(9a2  +  52). 

5.  {a}^  +  ¥')-^{a''  +  h^).     10.    (729c«+l)-- (96-2+1). 

Note.  The  introduction  of  negative  numbers  requires  an  exten- 
sion of  the  meanings  of  some  terms  common  to  arithmetic  and 
algebra.  But  every  such  extension  of  meaning  must  be  consistent 
with  the  sense  previously  attached  to  the  term  and  with  general  laws 
already  established. 

Addition  in  algebra  does  not  necessarily  imply  augmentation,  as 
it  does  in  arithmetic.  Thus,  7  +  (—  5)  =  2.  The  word  sum,  however, 
IS  used  to  denote  the  result. 

Such  a  result  is  called  the  algebraic  smn,  when  it  is  necessary 
to  distinguish  it  from  the  arithmetical  su7n,  which  would  be  obtained 
by  adding  the  absolute  values  of  the  numbers. 

The  general  definition  of  Addition  is,  the  operation  of  uniting 
two  or  more  numbers  in  a  single  expression  written  in  its  simplest 
form. 

The  general  definition  of  Subtraction  is,  the  operation  of  finding 
from  two  given  numbers,  called  minuend  and  subtrahend,  a  third 
number,  called  difference,  which  added  to  the  subtrahend  will  give 
the  minuend. 

The  general  definition  of  Multiplication  is,  the  operation  of  find- 
ing from  two  given  numbers,  called  multiplicand  and  multiplier,  a 
third  number,  called  product,  which  may  be  formed  from  the  mul- 
tiplicand as  the  multiplier  is  formed  from  unity. 

The  general  "definition  of  Division  is,  the  operation  of  finding 
the  other  factor  when  the  product  of  two  factors  and  one  factor  are 
given.  -, 

■•y 


^ 


CHAPTER  V. 
Simple  Equations. 


106.  An  equation  is  a  statement  that  two  expressions  are 
equal.     Thus,  4.r-12  =  8. 

107.  Every  equation  consists  of  two  parts,  called  the 
first  and  second  sides,  or  members,  of  the  equation. 

108.  An  identical  equation  is  one  in  which  the  two  sides 
are  equal,  whatever  numbers  the  letters  stand  for.  Thus, 
(x-{-b)(x-b)  =  :r^-h'. 

109.  An  equation  of  condition  is  one  which  is  true  only 
when  the  letters  stand  for  particular  values.  Thus,  a:  +  5 
=  8  is  true  only  when  x  =  S. 

110.  A  letter  ta  which  a  particular  value  must  be  given 
in  order  that  the  statement  contained  in  an  equation  may 
be  true  is  called  an  unknown  quantity. 

111.  The  value  of  the  unknown  quantity  is  the  number 
which  substituted  for  it  will  satisfy  the  equation,  and  is 
called  a  root  of  the  equation. 

112.  To  solve  an  equation  is  to  find  the  val|^  of  the 
unknown  quantity. 

113.  A  simple  equation  is  one  which  contains  only  the 
first  power  of  the  unknown  quantity,  and  is  also  called  an 
equation  of  the  first  degree. 


60  ALGEBRA. 


114.  Ij  equal  changes  he  made  in  both  sides  of  an  equa- 
tion^ the  results  will  be  equal.  §  43. 

(1)  To  find  the  value  oi xm  x-\-b^a. 

Subtract  h  from  each  side,     x-{-  b—b  =  a—b; 
Cancel +  &  — 6,  x  =  a—b. 

(2)  To  find  tlie  value  of  x  in  x  —  h  =  a. 

x  —  b  =  a; 
Subtract  —  b  from  each  side,  x—b  +  b  =  a  +  b; 
Cancel  — &+6,  x=  a  +  b. 

The  result  m  each  case  is  the  same  as  if  b  were  trans- 
posed to  the  other  side  of  the  equation  with  its  sign 
changed.     Therefore, 

115.  Ani/  term  may  be  transposed  from  one  side  of  an 
equation  to  the  other  provided  its  sign  be  changed. 

For,  in  this  transposition,  the  same  number  is  subtracted 
from  each  side  of  the  equation. 

116.  The  signs  of  all  the  terms  on  each  side  of  an  equa- 
tion may  be  changed ;  for,  this  is  in  effect  transposing  every 
term. 

117.  When  the  known  and  unknown  quantities  of  an 
equation  are  connected  by  the  sign  -f  or  — ,  they  may  be 
separated  by  transposing  the  known  quantities  to  one  side 
and  the  unknown  to  the  other. 

118.  Hence,  to  solve  an  equation  with  one  unknown 
quantity. 

Transpose  all  the  terms  involving  the  unknown  quantity 
to  the  left  side,  and  all  the  other  tertns  to  the  right  side: 


SIMPLE   EQUATIONS.  61 


combine  the  like  terms,  and  divide  both  sides  by  the  coeffidemi 
of  the  unknown  quantity. 

119.    To  verify  tlie  result,  substitute  the  value  of  the 
unknown  quantity  in  the  original  equation. 

Exercise  XXVIII. 

Find  the  value  of  x  in 

1.  5a:- 1  =  19.  8.  16a;-ll  =  7:r+70. 

2.  3a:  +  6=12.  9.  24a: - 49  =  19a: ~  14. 

3.  24a:  =  7a: +  34.  10.  3a:  +  23  =  78- 2a:. 

4.  8a:  — 29  =  26  — 3a:.  11.  26  — 8a:  =  80- 14ar. 

5.  12 -5a:  =  19 -12a:.  12.  13  -  3a:  =  5a:  — 3. 

6.  3a:  +  6-2a:=7a:.  13.  3a:- 22  =  7a:  +  6. 

7.  5a: +  50  =  4a: +  56.  14.  8  +  4a:  =  12a:- 16. 

15.  5a: -(3a: -7)  =  4a: -(6a: -35). 

16.  6a:-  2  (9  -4a:)  +  3  (5a:  -  7)  =  10a:-  (4  +  16a:  +  35). 

17.  9a:-3(5a:-6)  +  30  =  0. 

18.  a:  -  7  (4a:  -  11)  =  14  (a:  -  5)  --  19  (8  -  a:)  -  61. 

19.  (a:+7)(a:-3)  =  (a:-5)(a:-15). 

20.  (a;-8)(a:  +  12)  =  (a:  +  l)(a:-6). 

21.  (a:-2)(7-a:)  +  (a:-5)(a:  +  3)-2(a:-l)  +  12  =  0. 

22.  (2a: -7)  (a: +  5)  =  (9 -2a:)  (4 -a:) +  229. 

23.  14-a:-5(a:-3)(a:  +  2)  +  (5-:r)(4-5a:)=45a:-76. 

24.  {x-\-bf~{^~xf  =  2lx. 

25.  5  (a:  -  2)H  7  {x  -  2>f  =  (3a.-  -  7)  (4a:  -  19)  +  42. 


62  ALGEBRA. 


Exercise  XXIX. 

PROBLEMS. 

1.  Find  a   number  such  that  when    12  is  added  to  its 

double  the  sum  shall  be  28. 

Let  X  =  the  number. 

Then  2  a;  =  its  double, 

and        2a;  +  12  =  double  the  number  increased  by  12. 
But  28  =  double  the  number  increased  by  12. 

.'.2a;+ 12=  28. 

2a;  =28 -12, 
2a;  =16, 
a-=    8. 

2.  A  farmer  had  two  flocks  of  sheep,  each  containing  the 

same  number.  He  sold  21  sheep  from  one  flock  and 
70  from  the  other,  and  then  found  that  he  had  left  in 
one  flock  twice  as  many  as  in  the  other.  How  many 
had  he  in  each  ? 

Let  X  =  number  of  sheep  in  each  flock. 

Then  ar  —  21  =  number  of  sheep  left  in  one  flock, 
and         a;  —  70  =  number  of  sheep  left  in  the  other. 
.  • .  a;  -  21  =  2  (a;  -  70), 
a; -21=  2a; -140. 
a; -2a;  =  -140 +21, 
-    a;  =  -119, 
x=      119. 

3.  A  and  B  had  equal  sums  of  money ;  B  gave  A  $  5,  and 

then  3  times  A's  money  was  equal  to  11  times  B's 
money.    What  had  each  at  first  ? 

Let  X  =  number  of  dollars  each  had. 

Then        a;  +  5  =  number  of  dollars  A  had  after  receiving  $  5 
from  B, 
and  x  —  5  =  number  of  dollars  B  had  after  giving  A  |5. 


SIMPLE   EQUATIONS.  63 

.•.3(a;+5)=ll(a;-5), 
3x+15=lla;-55. 
3a:-llar=-55-15.  , 
-8a;  =-70, 
X  =  8f . 

Therefore,  each  had  $8.75. 

4.    Find  a  number  whose  treble  exceeds  50  by  as  much  as 
its  double  falls  short  of  40. 
Let  X  =  the  number. 

Then  3  a;  =  its  treble, 

and        3  05  —  50  =  the  excess  of  its  treble  over  50 ; 
also,       40  —  2  a;  =  the  number  its  double  lacks  of  40. 
.•.3a; -50  =  40 -2a;; 
3a;  +  2a;  =40 +  50. 
5a;  =90, 
a;  =18. 

6.   What  two  numbers  are  those  whose  difference  is  14, 
and  whose  sum  is  48  ? 
Let  X  =  the  larger  number. 

Then  48  —  a;  =  the  smaller  number, 

and        a;  —  (48  —  x)  =  the  difference  of  the  numbers. 
But  14  =  the  difference  of  the  numbers. 

-    .  .-.  a;-(48-a;)=14. 

a;  -  48  +  a;  =  14  ; 
2a;  =62; 

.T=3l. 

Therefore,  the  two  numbers  are  31  and  17. 

6.  To  the  double  of  a  certain  number  I  add  14,  and  obtain 

as  a  result  154.     What  is  the  number  ? 

7.  To  four  times  a  certain  number  I  add  16,  and  obtain  as 

a  result  188.     What  is  the  number  ? 

8.  By  adding  46  to  a  certain  number,  I  obtain  as  a  result 

a  number  three  times  as  large  as  the  original  number. 
Find  the  original  number. 


64  ALGEBRA. 


9.  One  number  is  three  times  as  large  as  another.  If  I 
take  the  smaller  from  16  and  the  greater  from  30, 
the  remainders  are  equal.     What  are  the  numbers  ? 

10.  Divide  the  number  92  into  four  parts,  such  that  the 

first  exceeds  the  second  by  10,  the  third  by  18,  and 
the  fourth  by  24. 

11.  The  sum  of  two  numbers  is  20 ;  and  if  three  times  the 

smaller  number  be  added  to  five  times  the  greater, 
the  sum  is  84.     What  are  the  numbers? 

12.  The  joint  ages  of  a  father  and  son  are  80  years.    If  the 

age  of  the  son  were  doubled,  he  would  be  10  years 
older  than  his  father.     What  is  the  age  of  each  ? 

13.  A  man  has  6  sons,  each  4  years  older  than  the  next 

younger.  The  eldest  is  three  times  as  old  as  the 
youngest.     What  is  the  age  of  each  ? 

14.  Add  $24  to  a  certain  sum  and  th^  amount  will  be  as 

much  above  $80  as  the  sum  is  below  $80.  What  is 
the  sum  ? 

15  Thirty  yards  of  cloth  and  40  yards  of  silk  together  cost 
$  330 ;  and  the  silk  cost  twice  as  much  a  yard  as  the 
cloth.     How  much  does  each  cost  a  yard  ? 

16.  Find  the  number  whose  double  increased  by  24  exceeds 

80  by  as  much  as  the  number  itself  is  less  than  100. 

17.  The  sum  of  $500  is  divided  among  A,  B,  0,  and  D.    A 

and  B  have  together  $280,  A  and  0  $260,  and  A  and 
D  $  220.     How  much  does  each  receive  ? 

18.  In  a  company  of  266  persons  composed  of  men,  women, 

and  children,  there  are  twice  as  many  men  as  women, 
and  twice  as  many  women  as  children.  How  many 
are  there  of  each  ? 


SIMPLE    EQUATIONS.  65 


19.  Find  two  numbers  differing  by  8,  such  tbat  four  times 

the  less  may  exceed  twice  tlie  greater  by  10. 

20.  A  is  58  years  older  than  B,  and  A's  age  is  as  much 

above  60  as  B's  age  is  below  50.  Find  the  age  of 
eacb. 

21.  A  man  leaves  bis  property,  amounting  to  ?  7500,  to  bo 

divided  among  bis  wife,  bis  two  sons,  and  tbree  daugh- 
ters, as  follows :  a  son  is  to  bave  twice  as  mucb  as  a 
daughter,  and  the  wife  $500  mofer  than  all  the  chil- 
dren together.     How  much  was  the  share  of  each  ? 

22.  A  vessel  containing  some  water  was  filled  by  pouring 

in  42  gallons,  and  there  was  then  in  the  vessel  seven 
times  as  much  as  at  first.  How  much  did  the  vessel 
hold? 

23.  A  has  $  72  and  B  has  $  52.     B  gives  A  a  certain  sum  ; 

then  A  has  three  times  as  much  as  B.  How  much 
did  A  receive  from  B  ? 

24.  Divide  90  into  two  such  parts  that  four  times  one  part 

may  be  equal  to  five  times  the  other. 

25.  Divide  60  into  two  such  parts  that  one  part  exceeds 

the  other  by  24. 

26.  Divide  84  into  two  such  parts  that  one  part  may  be  less 

than  the  other  by  36. 

Note  I.  When  we  have  to  compare  the  ages  of  two  persons  at  a 
given  time,  and  also  a  number  of  years  after  or  before  the  given 
time,  we  must  remember  that  both  persons  will  be  so  many  years 
older  or  younger. 

Thus,  if  a;  represent  A's  age,  and  2x  B's  age,  at  the  present  time, 
A's  age  five  years  ago  will  be  represented  by  x  —  b\  and  B's  by 
2.r  —  5.  A's  age  five  years  hence  will  be  represented  by  a;  +  5 ;  and 
B's  age  by  2  a;  +  5. 


66 


ALGEBRA. 


27.  A  is  twice  as  old  as  B,  and  22  years  ago  he  was  three 

times  as  old  as  B.     What  is  A's  age  ? 

28.  A  father  is  30  and  his  son  6  years  old.     In  how  many 

years  will  the  father  be  just  twice  as  old  as  the  son  ? 

29.  A  is  twice  as  old  as  B,  and  20  years  since  he  was  three 

times  as  old.     What  is  B's  age  ? 

30.  A  is  three  times  as  old  as  B,  and  19  years  hence  he 

will  be  only  twice  as  old  as  B.  What  is  the  age 
of  each  ? 

31.  A  man  has  three  nephews ;  his  age  is  50,  and  the  joint 

ages  of  the  nephews  is  42.  How  long  will  it  be  be- 
fore the  joint  ages  of  the  nephews  will  be  equal  to 
that  of  the  uncle  ? 

Note  II.  In  problems  involving  quantities  of  the  same  kind 
expressed  in  different  units,  we  must  be  careful  to  reduce  all  the  quan- 
tities to  the  same  unit. 

Thus,  if  X  denote  a  number  of  inches,  all  the  quantities  of  the  same 
kind  involved  in  the  problem  must  be  reduced  to  inches. 

32.  A  sum  of  money  consists  of  dollars  and  twenty-five-cent 

pieces,  and  amounts  to  $20.  The  number  of  coins  is 
50.     How  many  are  there  of  each  sort  ? 

33.  A  person  bought  30  pounds  of  sugar  of  two  different 

kinds,  and  paid  for  the  whole  $2.94.  The  better 
kind  cost  10  cents  a  pound  and  the  poorer  kind  7 
cents  a  pound.  How  many  pounds  were  there  of 
each  kind  ? 

34.  A  workman  was  hired  for  40  days,  at  $  1  for  every  day 

he  worked,  but  with  the  condition  that  for  every 
day  he  did  not  work  he  was  to  pay  45  cents  for  his 
board.  At  the  end  of  the  time  he  received  §22.60. 
How  many  days  did  he  work  ? 


SIMPLE    EQUATIONS.  67 

35.  A  wine  merchant  has  two  kinds  of  wine  ;  one  worth  50 

cents  a  quart,  and  the  other  75  cents  a  quart.  From 
these  he  wishes  to  make  a  mixture  of  100  gallons, 
worth  $2.40  a  gallon.  How  many  gallons  must  he 
take  of  each  kind. 

36.  A  gentleman  gave  some  children  10  cents  each,  and 

had  a  dollar  left.  He  found  that  he  would  have  re- 
quired one  dollar  more  to  enable  him  to  give  them 
15  cents  each.     How  many  children  were  there  ? 

37.  Two  casks  contain  equal  quantities  of  vinegar;  from 

the  first  cask  34  quarts  are  drawn,  from  the  second, 
20  gallons ;  the  quantity  remaining  in  one  vessel  is 
now  twice  that  in  the  other.  How  much  did  each 
cask  contain  at  first  ? 

38.  A  gentleman  hired  a  man  for  12  months,  at  theT  wages 

of  $  90  and  a  suit  of  clothes.  At  the  end  of  7  months 
the  man  quits  his  service  and  receives  $33.75  and 
the  suit  of  clothes.  What  was  the  price  of  the  suit 
of  clothes  ? 

—  39.    A  man  has  three  times  as  many  quarters  as  half-dollars, 
four  times  as  many  dimes  as  quarters,  and  twice  as 
many  half-dimes  as  dimes.     The  whole  sum  is  $  7.30. 
'    How  many  coins  has  he  altogether  ? 

40.  A  person  paid  a  bill  of  $15.25  with  quarters  and  half- 

dollars,  and  gave  51  pieces  of  money  altogether. 
How  many  of  each  kind  were  there  ? 

41.  A  bill  of  100  pounds  was  paid  with  guineas  (21  shil- 

lings) and  half-crowns  (22-  shillings),  and  48  more 
half-crowns  than  guineas  were  used.  How  many  of 
each  were  paid  ? 


CHAPTER  VI. 

^  Factors. 

120.  In  multiplication  we  determine  the  -product  of  two 
given  factors ;  it  is  often  important  to  determine  \hQ  factors 
of  a  given  product. 

121.  Case  I.  The  simplest  case  is  that  in  which  all  the 
terms  of  an  expression  have  one  common  factor.     Thus, 

(1)  :i?  -^  xy  ^^  X  {x -\-  y). 

(2)  6a3  +  4a2  +  8a  =  2a(3a2  +  2a  +  4). 

(3)  l^a?h  -  21a^h''  +  36a6  =  9a^  (2^2  _  3a5  +  4). 

Exercise  XXX. 
Resolve  into  factors : 

1.  5a2-15a.  4.   4.oi^y -l2xY +  ^xf. 

2.  6a^+18a2-12a.  5.    y^ -  mf  ^- by\-\- cy . 

3.  49ar^-21:r+14.  6.    (ja'b^ -2\a'h'' +  21o^h\ 

7.  54  .lY  4- 108  xy  -  243  x^- 

8.  45:ry»-90a;^/-360.Ty- 

9.  70  aV  -  140  ah/  +  210  aif. 
10.   32  a%^  +  96  a«5«  -  128  a^b''. 


FACTORS.  69 


122.    Case  II.    Frequently  the  terms  of   an  expression 
can  be  so  arranged  as  to  show  a  common  factor.     Thus, 

(1)  x^-{-ax-\-  hx  +  (2^  =  (^  +  ax)  +  (hx  +  ah), 

=^  X  (x  -\-  a)  -\-  h  (x  -\-  a), 
=  {x  +  b){x-{-a). 

(2)  ac— ad~hc-\-hd^{ac  —  ad)  —  {hc  —  hd), 

=  a(c  —  d)  —  b  (c  -d), 
=  {a-h){c-d). 


Exercise  XXXI. 
Resolve  into  factors  :  * 

1.  x^  —  ax  —  hx-\-  ah.  6.   ahx  —  ahy  -\-pqx  —  pqy. 

2.  ab  -{-ay  —  by  —  if.  7.    cdy?  -\-  adxy  —  bcxy  —  aby'^. 

3.  he Arbx—  ex  —  0?.  8.    ahoy  —  b^dy—  acdx  +  bd^x. 

4.  vtx  -f-  "rnn  +  ax  +  aw.  9.    ax  —  ay  —  bx-{-  by. 

5.  cdoc^  —  cxy  +  cfey  —  %^.  10.    c^c^z^  —  cyz  +  (iyz  —  y^. 

123.  The  square  root  of  a  number  is  one  of  the  two  equal 
factors  of  that  number.  Thus,  the  square  root  of  25  is  5 ; 
for,  25  =  5X5. 

The  square  root  of  d^  is  o? ;  for,  a^  =  a^  X  a^. 

The  square  root  of  a^^V  is  ahc ;  for,  a^5V  =  ahc  X  abe. 

In  general,  the  square  root  of  a  power  of  a  number  is  ex- 
pressed by  writing  the  number  with  an  exponent  equal  to 
one-half  the  exponent  of  the  power. 

The  square  root  of  a  product  may  be  found  by  taking 
the  square  root  of  each  factor,  and  finding  the  product  of 
the  roots. 


70  ALGEBRA. 


The  square  root  of  a  positive  number  may  be  either  posi- 
tive or  negative ;  for, 

or,  a^  =  —  a  X  —  a ; 

but  throughout  this  chapter  only  the  positive  value  of  the 
square  root  will  be  taken. 

124.  Case  III.  From  §  73  it  is  seen  that  a  trinomial 
is  often  the  product  of  two  binomials.  Conversely,  a  trino- 
mial may,  in  certain  cases,  be  resolved  into  two  binomial 
factors.     Thus, 

To  find  the  factors  of 

x2+7:r+12. 

The  trst  term  of  each  binomial  factor  will  obviously  be  x. 

The  second  terms  of  the  two  binomial  factors  must  be  two 
numbers 

whose  product  is  12, 
and  whose  sum  is  7. 

The  only  two  numbers  whose  product  is  12  and  whose 
sum  is  7  are  4  and  3. 

Again,  to  find  the  factors  oi  a?  -\- b  xy  -\- Q 'if . 

The  first  term  of  each  binomial  factor  will  obviously  be  x. 

The  second  terms  of  the  two  binomial  factors  must  be  two 

numbers 

vfho&e  product  is  6?/^, 
and  whose  sum  is        5y. 

The  only  two  numbers  whose  product  is  6y^  and  whose 
sum  is  Sy  are  3?/  and  2?/. 

.-.  x'  -i-bxy +  &,/-=  (x  +  3i/)  (x  +  22/). 


FACTORS.  7i 


Exercise  XXXII. 

Find  the  factors  of  : 

1.  rr2+ll^  +  24.  11.  x^  +  lSax-i-SQc^, 

2.  ci^+llx  +  SO.  12.  y" -{•  19 pi/ +  ^8 f. 

3.  f  +  172/  +  60.  13.  224-29^3  +  100^. 

4.  2^  +  132  +  12.  14.  a'  +  ba^  +  Q. 

5.  a:2_|_9i^_|_iio.  15.  2«  +  423  +  3. 

6.  ?/2_^35y4_300.  1&  a252^i8a6  +  32. 

7.  62  +  236  +  102!     •  17.  a:«y*+7a;y+12, 
a  a:2  +  3a7  +  2.  la  z^o+lOz^  +  lG. 
9.  a:2+7ar  +  6.  19.  a'+9ab  +  20b^. 

10.  a2  +  9a6  +  862.  20.   a;«  +  9 ^z;^  +  20. 

21.  aV+14a6a;  +  3362.  24.  6V  +  18  oic  +  65  a^. 

22.  a^c^ +7acx -i- 10  x^.  25.  r^s^ -}_  23  rsz  +  90  z^. 

23.  a;2^22_^  19^^2  +  48.  26.  /mV  +  20  mVj9^  +  51/g*. 

125.   Case  IV.    To  find  the  factors  of 

^-907  +  20. 

The  second  terms  of  the  two  binomial  factors  must  be  two 
numbers 

whose  product  is  20, 
and  whose  sum  is      —  9, 

The  only  two  numbers  whose  product  is  20  and  whose 
sum  is  —  9  are  —  5  and  —  4. 

.'.  x"  ~9x  +  20  =  (x  -  5)(x  -  4:). 


72  ALGEBRA. 


Exercise  XXXIII. 
Resolve  into  factors : 

1.  ^-7a:  +  10.  13.   a2^V-13a5c  +  22. 

2.  :r2-29:r  +  190^  14.   ar'-l^x  +  bO. 

3.  a^- 23a +  132.  15.   a^-20x  +  100. 

4.  b^-SOb  +  200.  16.   aV-21a:r  +  54.  /^     3 

5.  2»-43z  +  460.'  17.   aV- 16a5a:  +  39//.       ^ 

6.  5:2-7^  +  6.  la   aV - 24 acz  + 14322. 

7.  a;*-4aV  +  3a^  19.   a:2-i_20a;  + 91. 

8.  x'-Sx+U.  20.   a;2_23a7+120. 

9.  2^  ~  biz +  56.  21.   22_532_|.36o. 

10.  ^/^-Ty'+ia  22.   aP-(a+c)x-\-ac. 

11.  :r2i/2_27a;y  +  26.  23.   y2/_  28  a^»y2  + 187^2^2 

12.  a'b^~lla^b^  +  m.  24.   c2J2_  3o^5^j_|_221a262. 


y^^/J^ 


126.    CaseV.    To  find  the  factors  of 
^+2^-3. 

The  second  terms  of  the  two  binomial  factors  must  be 

two  numbers 

whose  product  is  —  3, 

and  whose  sum  is       +  2. 

The  only  two  numbers  whose  product  is  —  3  and  whose 
sum  is  +  2  are  +  3  and  —  1. 

.'.  :>?+2x~Z  =  {x  +  ^){x-\). 


FACTORS.  73 


Exercise  XXXIV. 
Resolve  into  factors : 

1.  c^-\-Qx-1.  8.  a2  +  25a-150. 

2.  a^  +  bx-M.  9.  ^'«  +  35*-4. 

3.  3/2+7y-60.  10.  6V  +  35c-154. 

4.  y2_pi2y_45.  11.  c-^«  +  15c*-100. 

5.  z2^  112-12.  12.  6-2  +  17^-390. 

6.  22_|.i32_i4o.  13.  a? -\.  a -1^2. 

7.  a2_^13a-300.  14.  a^ fz^  +  ^ xyz -  22. 

127.    Case  VI.   To  find  the  factors  of 
3^-bx-m. 

The  second  terms  of  the  two  binomial  factors  must  be 

two  nurabers  i  j    ^  •        en 

whose  product  is  —  do, 

and  whose  sum  is        —    5. 

The  only  two  numbers  whose  product  is  —  66  and  whose 
sum  is  —  5  are  —  11  and  +  6. 

.\x^-bx~m  =  (x  -11)  (x  +  6). 

Exercise  XXXV. 
Resolve  into  factors : 

1.  ar^-Sx-2S.  6.  a^- 15a- 100. 

2.  y'-7y-18.  7.  c-i«-9^-10. 

3.  ^^-9a;-36.  8.  ^-8:i;-20. 

4.  22_ii2_60.  9.  f-ba7/~50a\ 

5.  z2-132;-14.  10.  aW-Sab-4.. 


74  ALGEBRA. 


11.  aV-3aa;-54.  14.   ?/V-5yV-84. 

12.  c2cZ2-24cc^-180.         15.    «2^2_]^(3^5_3g^ 

13.  aV-a3c-2.  16.    :?;2  _  (^  _  j)  ^  _  ^^5 

We  now  proceedjo  the  consideration  of  trinomials  which 
are  perfect  squares^  These  are  only  particular  forms  of 
Cases  III.  and  IV.,  but  from  their  importance  demand 
special  attention. 

128.    Case  VII.    To  find  the  factors  of 

The  second  terms  of  the  two  binomial  factors  must  be 

two  numbers  n  ^    ^  •    oi 

whose  product  is  81, 

and  whose  sum  is        18. 

The  only  two  numbers  whose  product  is  81  and  whose 
sum  is  18  are  9  and  9. 

.-.  x^  +  18^'  +  81  =  (.r  +  9)  (x  -Jr9)  =  (x  +  df. 

Exercise  XXXVI. 
Resolve  into  factors : 

1.  x'+12x-i-S6.  8.   9/+l6fz^^64:z\ 

2.  x^-\-2Sx+19Q.  9.   /  + 242/3 +144. 

3.  :r2  +  34:r  +  289.  10.   xh""  +  162  xz  +  6561 . 

4.  22_|.22;+1.  11.    4:a'-{-12ab'--{-db\ 

5.  ^2  _|_200y  + 10,000.  12.    9^:2^^  +  30^2  +  2522. 

6.  z^+14z2  +  49.  13.    9:r2+12xy  +  4?/2. 

7.  rr2  +  36:r?/  +  324y2.  14.    4  a*^'^  _|_  20  aVy  +  25  .i-y. 


FACTORS.  75 


129.    Case  VIII.    To  find  the  factors  of 
x'-lSx  +  Sl. 

The  second  terms  of  the  two  trinomials  must  be  two 

numbers 

vfhose  product  is  81, 

and  whose  sum  is    —  18. 

The  only  two  numbers  whose  product  is  81  and  whose 
sum  is  —  18  are  —  9  and  —  9. 

.-.  sr"-  18:?;  4-  81  =  (x  -  9)  (x-9)  =  (x-9y. 

Exercise  XXXVII. 

1.  a^-8a-{-16.  10.  4a;y- 20:ryz  + 25?/V. 

2.  a^-SOa-i-225.  11.  IQx"/ -Sxfz^  +  fz*. 

3.  x'-SSxi-Sei.  12.  da^b^c'--6ah'c'd+hh^d\ 

4.  5;2_4o^-f-400.  13.  16a^ -8xy  +  ar^i/. 

5.  ?/2_  100?/  + 2500.  14.  a^x*-2a^bx^7/*i-by. 

6.  ?/^-20y2-j-100.  15.  SQa^if  -  60x7/-{-26p*. 

7.  y2„  50^2  +  62522.  16.  l~6ah^  +  9a^b\ 

8.  07^- 32:^:2^2  _!_  256  y.  17.  9mV-24mn  +  16. 

9.  z^-3423  +  289.  18.  ^Px" -12b x^ij +  90^7/". 

19.  49a2_li2a^»  +  64Z.2 

20.  64xy -  1605;y2  +  100 :rV. 

21.  49d'b'c'-28abcx  +  4:x'. 

22.  121a;^-286.r2y  +  1693/^. 

23.  289  x^i/z^  -  102  x7/z'd  +  9  ^/z^"^ . 

24.  361  xYz^  -  76  aSc^y^  +  4  a%-2. 


76  ALGEBRA. 


130.  Case  IX.  An  expression  in  the  form  of  two  squares, 
with  the  negative  sign  between  them,  is  the  product  of  two 
factors  which  may  be  determined  as  follows : 

Take  the  square  root  of  the  first  number,  and  the  square 
root  of  the  second  number. 

The  sum  of  these  roots  will  form  the  first  factor ; 

The  difference  of  these  roots  will  form  the  second  factor. 
Thus : 

(1)  o:'-b^  =  {a  +  b){a-h). 

(2)  a'-ih-cf=^\a+(b-c)\\a-{b-c)\, 

=  \a-\-b  —  c\  \a  —  b-\-c\. 

(3)  (a-bf-ic-df=\{a-b)-{-{c-d)\\{a-b)-{c-d)\, 

^^la—b-^-c  —  d]  \a  —  b—c-]-d\. 

131.  The  terms  of  an  expression  may  often  be  arranged 
so  as  to  form  two  squares  with  the  negative  sign  between 
them,  and  the  expression  can  then  be  resolved  into  factors. 
Thus: 

a^  -\-b^  -  c"  -  d^  -{-'lab  +  2cd, 

=  o?  +  2ab  -\-b^ -  c"  +  2cd-  d\ 

=  {a^-\-'lab-^W')-{(?-2cd-^d% 

=  {a^b)^-{c-d)\ 

=  \{a^b)-^{c-d)\\{a^-b)~{c-^d)\, 

-=\a^b^c-d\\a-\-b-c-\-d\. 

132.  An  expression  may  often  be  resolved  into  three  or 
more  factors.     Thus : 

(1)    x"-^  -f'^^i^:^  ^if){a?  -if) 

=  (^  +  /)  {^*  +  y')  i^^  +  f)  if-f) 

-=^{o?-\-f){x''-\-y'){^:^^y'){x-^y){x~-y). 


FACTORS.  77 


(2)   ^{ab-\-cd)^~{a^-\-h^-(?-d^J, 

=  \2ah  +  2cd  +  a^  +  b^-  c'  -  d^ 

\2ab  +  2cd-a^-b^  +  c^+-d''\, 
=  \(a^  +  2ab  +  b^'j-^e"  -  2cd  +  d'')l 

K^  i-2cd+  d^)  -  {a?  -  2ab  +  b^)\, 
^\{a  +  bf-{o-df\\{c  +  dr-{a-bf\, 
=  \a-\'b  +  {c-d)\\a-\-b-{c-d)] 

\e  +  d+{a-b)\\c-\-d-{a-b)\, 
^\a  +  b-\-c-d\\a  +  b-c^d\ 

\c  +  d-ya~-b\\c  +  d-a  +  b\. 

Exercise  XXXVIII. 
Resolve  into  factors : 

1.  a^-b\  14.  {a-\-by~{c  +  d)\ 

2.  a^-ie.  15.  {x  +  yf-{x-yf. 

3.  4a2_25.  la  2ab-a^'--W+l. 

4.  a^-b\  17.  3?  -2yz-f-^. 

5.  a^-1.  18.  ^-2^y  +  y2_^ 

6.  a«-58.  19.  a^-\-12bc~^b''~^c'. 

7.  a»-l.  20.  a^~2ay^'if~o?-2xz-^. 
a    36r^-49y2.  21.  2:ry-:r-y2_|_,2_ 

9.  Vd^x'f~\2\a%\  22.  :r2  4_^_^22-- J2__2a;y-2c/2. 

10.  1-49^:2.  23.  ^-y2  +  z2_^2_2^^_j.2^^^ 

11.  a^-2552  24.  2a5  +  a2_|-52__^^ 

12.  (a-^j^-c-^.  25.  2xy-x^-t^-\-d?-\-b^~2ah, 

13.  :r2-(a-^)^  26.  (aa;  +  5y/-l. 


'°  ALGEBRA. 


27.  \~a?~f-\.<2.xy.  31.    {x -^  Xf  -  {y -  Yf 

28.  (5«-2)2-(a-_4)^.  32.    d^-^^^^y^^^ 

29.  a2-2aZ»  +  ^.2_^2  33^    a^- ^2_  ^^.^  -  ^2 
(^+l)^-(y  +  l/.  34.    4^^-9a;«+6:r-'l 


# 


133.   Case  X. 

^"^  T^-^^^  +  A  +  ^y^  +  ^y^+y, 

and  so  on,  it  follows  that  the  difference  between  two  equal 
odd  powers  of  two  numbers  is  divisible  by  the  difference 
between  the  numbers. 


Exercise  XXXIX. 
Eesolve  into  factors : 

1-  «'-^'-  6.  8a7«-272/«. 

^-  ^-^-  7.  643/«-1000/. 

3.  a;3-343.  8.  729:^3-512^. 

4.  ?,'3-125.  9.  27a3-i728. 

5.  y«-216.  10.  lOOOaS- 1331^3 

134.    Case  XI. 

bince  — , =  ar  —  r/.-r  +  a^ 

and  so  on,  it  follows  that  the  sum  of  two  equal  odd  powers 
of  two  numbers  is  divisible  by  the  sum  of  the  numbers. 


i^^J 


FACTORS.  79 

EXEECISE   XL. 

Resolve  into  factors : 

1.  ar'  +  f.  6.  216a»  +  512c3. 

2.  a^  +  8.  7.  729^:3+17282/3^ 

3.  :^+216.  8.  a^  +  y^. 

4.  f  +  Q4:2^.  9.  x^  +  yl 

5.  64  Z'«+ 125^3.  10.  32i^  +  243c«. 

135.  Case  XII.  The  sum  of  any  two  powers  of  two 
numbers,  whose  exponents  contain  the  same  odd  factor,  is 
divisible  by  the  sum  of  the  powers  obtained  by  dividing 
the  exponents  of  the  given  powers  by  this  odd  factor. 

Thus, 

In  like  manner,  x"'-i-S2y^,  which  is  equal  to  rc'°  +  (2?/)^ 
is  divisible  by  :r'  +  2?/;  but  x*-}-y\  whose  exponents  do  not 
contain  an  odd  factor,  and  x^-{-y^°,  whose  exponents  do 
not  contain  the  same  odd  factor,  cannot  be  resolved  into 
factors. 


Exercise  XLI. 
Resolve  into  factors : 

1.  a«  +  6«.  3.    x'^  +  'i/''.         5.    rr^  +  l.  7.    64a«  +  :r«. 

2.  «!«  +  &«».       4.    b^  +  64:c^       6.    a^+1.         8.    729  +  c«. 


80  ALGEBRA. 


136.  Case  XIII.  For  a  trinomial  to  be  a  "perfect  square, 
the  middle  tei-vi  viust  be  twice  the  product  of  the  square 
roots  of  the  first  and  last  terTns. 

The  expression  x*  +  s^i/  +  y*  will  become  a  perfect 
square  if  s^}/^  be  added  to  the  middle  term.  And  if  the 
subtraction  of  a^'if  from  the  expression  thus  obtained  be  in- 
dicated, the  result  will  be  the  difference  of  two  squares. 
Thus, 

=  {x^  +  f  -{- xy){x^  -^  f  -  xy), 
OT,{x^  +  xy-]-y^){f-xy-\-f). 

Exercise  XLII. 
Resolve  into  factors  • 

1.  a^  +  a^b^-\-h\  8.   49m^+ 110wV  +  81n^ 

2.  9a;*  +  3a;2^  +  4y^  9.    9a^-f21aV  + 25e^ 

3.  I^x''-l1x'f  +  y\  10.   ^9a^-lbaW-\~l2lb\ 

4.  81a^  +  23a262  +  16^^  11.   64^^*+ 128:1^  +  81?/^ 

5.  81a^-28a2^2^16  5^  12.*4a;^  -  37:zr2/4- g^^ 

6.  92;*  +  38a;2^  +  492/^  ^13.   25^^-41^,^+ 16y^ 

7.  25  a*  -  9^252+ 16  ^^  14.   81a;^  -  34a,y +  y*. 

*  If,  in  Example  12,  9y*=  (-  ?>y^f,  then  25 .r*/  should  be  added  to 
4ic*  —  Zla?y^  +  93/^,  in  order  to  make  the  expression  a  perfect  square 
That  is,  we  should  have : 

(4a;* -  12.r^/  +  dxf)  -  2d:^y\ 
=  {2a»-Sy^f-25x'f, 
=  (2.-^2-3/  +  5xy){2x^  -  3y«  -  5xy), 
or,  (2a-«  +  5xy-  Sy'^){2x'  -5xy-  3/). 


^ 


FACTORS.  81 


I     i- 


137.    Case  XIV.    To  find  the  factors  of 
Gx'  +  x-U. 

It  is  evident  that  the  first  terms  of  the  two  factors  might 
be  6  a;  and  x,  or  2x  and  Sx,  since  the  product  of  either  of 
these  pairs  is  6x^. 

Likewise,  the  last  terms  of  the  two  factors  might  be  12 
and  I,  6  and  2,  or  4  and  3  (if  we  disregard  the  signs). 

From  these  it  is  necessary  to  select  such  as  will  produce 
the  middle  term  of  the  trinomial.  And  they  are  found  by 
trial  to  be  3  a;  and  2x,  and  —  4  and  +  3. 

/.  6x'  +  x-12  =  (Sx-4:)(2x-{-^). 

Exercise  XLIII. 
Resolve  into  factors : 

1.  12r^-5a;-2.  13.  GaV  +  ox-l. 

2.  Uar'-lx+l.  14.  6b^~1bx-3a^. 

3.  Vl3?-x—\.  15.  4a72  +  8a;  +  3. 

4.  ?>oi?-2x-^.  16.  a^-ax-^x?. 

5.  Zx'-\-^x-^.  17.  8a2+I4a^>-I5^2 

6.  60.-2-}- 5a; -4.  18.  6ci2-19ac  +  10£?2. 

7.  ^x^-\-\?>x-\-Z.  19.  80.-2  + 34a;?/ +  21/. 

8.  4a.-2+lla;-3.  20.  '^x^ -22xy -2\%f. 

9.  4a;2__4^_3_^  21.  6a.-2+ 19a;y- 7y*. , 

10.  x^-Zax-\-2o?.  22.    lIa2-23a6  +  252. 

11.  12a^  +  aV-a;r  23.    2(? -XZcd-^^d}. 

12.  2:^-\-hxy-^2y'.  24.   6^2  +  7^2-321 


y-tA 


82  ALGEBRA. 


138.  Case  XV.  The  factors,  if  any  exist,  of  a  polyno- 
mial of  more  than  three  terms  can  often  be  found  by  the 
application  of  principles  already  explained.  Thus  it  is  seen 
at  a  glance  that  the  expression 

fulfils,  both  in  respect  to  exponents  and  coefficients,  the  laws 
stated  in  §  83  for  writing  the  power  of  a  binomial ;  and  it 
is  known  at  once  that 

Again,  it  is  seen  that  the  expression 

consists  of  three  squares  and  three  double  products,  and 
from  §  79,  is  the  square  of  a  trinomial  which  has  for  terms 
X,  y,  z. 

It  is  also  seen  from  the  double  product  —  2  :r?/,  that  x  and 
y  have  unlike  signs ; 

and  from  the  double  product  2xz,  that  x  and  z  have  like 
signs.     Hence, 

a^—2xy-\-'i^-\-2xz  —  2yz  -\-^  =  {x  —  y-\-  z/. 

Exercise  XLIV. 
Resolve  into  factors : 

1.  a^-f  3a25  +  3a52  +  ^>3.      4.    x^^4:C^y+Qx'f-^4.xf+y\ 

2.  a3  +  3a2  +  3a  +  l.  5.    x^  -  4::^ +^x^ -^x-\-l. 

3.  a»-3a2  4-3a-l.  6.    a^-4a^c  +  6aV-4a6'«+c^ 

7.  x^  +  2xy-{-7f  +  2xz-{-2yz  +  ^. 

8.  x^-2xy  +  i/-2xz-\-2yz  +  z^. 

9.  a^ -\-h^ -\- (^ ^2ab  -2ac ~2bc. 


FACTORS.  83 


139.    Case  XVI.    Multiply  2:r-y  +  3  by  a;  +  2y-3. 

2a;  -      y  +  3 
X  +    2y-3 
2o?—    xy  -{-Zx 

^xy-lf  +6y 

-6a:  +  3y-9 

■   2a:2_p3^^_2^_3^4_9y_9 

It  is  to  be  observed  that  2x^+3x2/  — 2 3/*^,  of  the  product,  is  obtained 
from  (2x  — y)  X  (x  +  2y)  ; 

that  —  9  is  obtained  from  3  x  —  3  ; 

that  —  3  X  is  the  sum  of  2x  x  —  3  and  x  X  3 ; 

that  9y  is  the  sum  of  2y  x  3  and  —  3/  x  —  3. 

From  this  result  may  be  deduced  a  method  of  resolving 
into  its  factors  a  polynomial  which  is  composed  of  two  tri- 
nomial factors.     Thus : 

Find  the  factors  of 

6a;2  _  7^y  _  3y2  _  9^  _|.  3Q^  _  27. 
The  factors  of  the  first  three  terms  are  (by  Case  XIV.) 
3x  +  y  and  2x  —  3y. 

Now  —27  must  be  resolved  into  two  factors  such  that  the  sum  of 
the  products  obtained  by  multiplying  one  of  these  factors  by  3x  and 
the  other  by  2x  shall  be  —  9x. 

These  two  factors  evidently  are  -  9  and  +  3. 

That  is,  (6^:2  _  7^^  _  3^  _  9^  _l_  30y  _  27) 

=  (3a;-f-y-9)(2a;-3y-f-3). 


140.    The  following  method  is  often  most  convenient  for 
separating  a  polynomial  into  its  factors  : 
Find  the  factors  of 

2:^-^xy-\-'lf-^  Ixz  -  byz  +  32^. 


1.  Reject  the  terms  that  contain  z. 

2.  Reject  the  terms  that  contain  y. 

3.  Reject  the  terms  that  contain  x. 


84  ALGEBRA. 


Factor  the  expression  that  remains  in  each  case. 

1.  23?-bxy^-2y^={x-2y){2x-y). 

2.  2x^  +  1xz^2>z'  =  {x  +  ^z){2x+z). 

3.  2/-53/2  +  322  =  (_2y  +  32)(-y+4 

Arrange  these  three  pairs  of  factors  in  two  rows  of  three  factors 
each,  so  that  any  two  factors  of  each  row  may  have  a  common  term. 
Thns: 

x-2y,  X  ^-2>z,  -2y +  Zz\ 

2x  —  y,  2x  +  z,  —y  +  z. 

From  the  first  row,  select  the  terms  common  to  two  factors  for 
one  trinomial  factor : 

x-2y  +  32. 

From  the  second  row,  select  the  terms  common  to  two  factors  for 
the  other  trinomial  factor : 

2x  —  y  +  z. 

Then,  2a^ -5xy  +  2y^-]-7xz  -  5^z  +  Sz^ 

=  (x-27/  +  3z)(i2x-7/  +  z). 

141.  When  a  factor  obtained  from  the  first  three  terms 
is  also  a  factor  of  the  remaining  terms,  the  expression  is 
easily  resolved.     Thus : 

(3)   x^-Sxi/  +  2f-^x  +  6i/, 
=  (^x-2y){x-y)-^{x-2yl 
=  {x-2y){x-y-^). 

Exercise  XLV.' 
Resolve  into  factors : 

1.  2:r2-5a:y  +  2y2-17:r-hl3y  +  21. 

2.  ^x^-Z1xy-\-Q>if  —  bx-'oy-l. 

3.  ^a^  —  bxy  —  ^y^  —  x  —  by—1. 

4.  bx'-^xy  +  2>f+1x-by-\-2. 


FACTORS.  85 


6.  x^-2bf-l0x-2Qy-\-2l. 

7.  2o^  —  bxy-\-2'i/^  —  xz  —  yz  —  7?. 

8.  ^x^  +  xy-f-Zxz-^^yz-'^z^. 

9.  ^x^—1xy-{-'i^-]-d>bxz  —  byz  —  ^7?. 

10.  bx^-d>xy-^2>f-2,xz  +  yz-2:?. 

11.  2a^-xy-Zf-byz-2^. 

12!   ^a?  —  l?>xy-\-^f-]-12xz  —  \2>yz-]-^z^, 

13.  a;2_2a:y4-2/2_|_5^_5^_ 

14.  2a:2_|_5^y_3^_4^2  +  2yz. 


Exercise  XL VI. 

MISCELLANEOUS    EXAMPLES. 

The  following  expressions  are  to  be  resolved  into  factors 
by  the  principles  already  explained.  The  student  should 
first  carefully  remove  all  monomial  factors  from  the  ex- 
pressions. 

1.  5a:2_i5^_20.  9.  a*  +  a'+l. 

2.  2o(^ -1^x^  +  24:3^.  10.  x"  -'i^-xz  +  yz. 

3.  3a'^2_9^5_i2.  til.  ah-ac-h^-\-hc. 

4.  a^-\-2ax-\-oi^-]-^a-\-4iX.  12.  ?>a? —  2>xz  —  xy +  yz. 

5.  a^-2ah-{-¥-c^.  13.  a^~x^-ah-hx. 

6.  :r2_2^y+y2_^^_^2^c?-c?-.    14.  a^  -  2 03;  +  a;2_^  ^ _ ^ 

7.  ■4-n'2_2:i^-a;^^  ■5.  Sx^  -  Sy^  _2a;  + 2y. 
«.  o?-h^-a~h.  16.  ^^  +  a;3  +  :e24-a;. 

17.  aV-aV-aV+1. 

18.  ^:^-2s^y~21xf-\-l^f. 


86 


ALGEBRA. 


A9.    4cx'~a^-{-2x-l.       28.    4:a' -  4:ah -^  b\ 

20.  x'~7/.  29.    IGT'-SOxiz+lOOf. 

21.  x^-i-f.  30.   36aV/-25Z^V/. 

22.  729 -a;«  31.    9  ^r^y^  -  30  ^ry^^  +  25  2^. 

23.  a;^y  +  y«.  32.    IQx'-x. 

24.  a^^-^T*.  33.    a^-2x7/-2xz+y'+27/z+z\ 

25.  :r2_|_4^_21.  34.   a' -  a6  -  6^>2  _  4«  _|_  125. 

26.  Sa'~21ab-}-S0b'.      35.    .^^  +  20:^ +  /-:r-y  -  6. 

27.  2x'-^x'i/-6a^f.      36.    (a  +  5)4__c-^ 

37.  .r--.Ty-6y2_4.^4_lo,,.     39.    3^- ll^ry  +  Gy^. 

38.  l-a7  +  5:2_^  ^^    a^-i-20x-^91. 
-  41.    (:r  -  y)  (:r2  -  2^)  _  (:,.  _  ,)  (^-^  __  ,f~^ 

42.  :r2_5^_24.  ^    y2_4^_117_ 

43.  {x^-f-^^J_^,f,.  51^    ^+6:r-135. 

44.  5a;«y2  +  5a:2^z-60a;2;2.        52.    ^a^  ~\2ab -\-9h'~  \(?. 

45.  3a:«-a:2^3^_^  ^^^    {a-YZhf -9  (h  ~  cf. 
40    a72-2ma:  +  7;?2_;^2  ^^    92^-4y2-f  4yz-z2 
47.    ^aW-{^a'-^V^-^)\         55.    6^V- 7^x^- S^-^ 
^V«'  +  ^*-                                     56.    a^-W-~?>ab{ci~h). 
49.    l~14a«^  +  49aV.             57.    x^  ^if -\-Zxy  {^x^-y). 

58.  a^  -  53  _  ^  ^^,2  _  ^2^  +  6  (a  _  ^,)2   ^ 

59.  ^x'f-Zxf-^Sf.  60.    6:t^  +  13^y  +  6/. 

61.  6^252  __  ^^3  _  12  54 

62.  a^  ^2ad^  d''  ~  ^b'  -YVlbc  ~9(^. 

63.  :r«-2a'2y  4-4^^-8/.      64.    \a\i' --^abx -^2>b\ 


FACTORS.  87 


65.  18.r2-24a:y-f-8/-^95;-6y.      74.  I6a^x-2x\ 

66.  2a^  +  2xy-l2f  +  6xz-i-l87/z.    75.  S2da^-4:bf, 

67.  (x  +  7/f-l-xi/(x-\-^  +  l).     /76.  a;-27x^ 
88.  x^-f-z^-\-2yz-{-x-\-^-z.       77.  x^-^f\ 

69.  2^  +  4:i:y  +  23/«  +  2aa:  +  2ay.      78.  49^^-121^2. 

70.  16  a2^  +  32a6c  + 126^2.  79.  16-81y*. 

71.  m^p  —  m^q  —  n^jj-^-yi^q,  80.  12z*  — 2*  — 6. 

72.  12a2r2-14aa;y-6a/.  81.  a;^- a;2  +  a;- 1. 

73.  2a^  +  4:0i?-1idx,  82.  a;2_^2a; -f- 1 -y*. 

83.  49  (a  -  ^)2  -  64  (m  -  n)^.  ' 

\       2ab      )' 

85.  o(^-b?>x  +  2,m, 

86.  5;^-2:c2y  +  a,^-4a:  +  8y-4. 

87.  2a6-26c-a6-f  ce  +  2Z>2--6e. 

88.  125a;*  +  350a;*3/2+245;ry*. 

X89.  a^  +  a-'^  +  ft^^'^^a'^Ha'^^  +  a^A 

90.  2  a*.r  —  2  a^cx  -f  2  ac^^  —  2  c*^;. 

91.  6:r*-5:cy-6?/2  +  3arz  +  15yz-92*. 

92.  ^x^-^xy-\-2f-2>xz-yz-:^. 
''93.  Sa'^-To^-f 25=»  +  5ac-56c?  +  2c*. 

^^  x*^  -  2a,-«  +  a;'-  8:t-  +  8. 

)( 96.    5x^  -  8:ry  +  3?/2  _  5:^:  -f  3y. 

96.  a^  _2rtc^4-c?2- 4^.2 -f-l25(?- 9^2. 

97.  (:c2_^_6)(^^_^_20). 


CHAPTER  VII. 
Common  Factors  and  Multiples. 

142.  A  common  factor  of  two  or  more  expressions  is  an 
expression  which  is  contained  in  each  of  them  without  a 
remainder.     Thus, 

5  (2  is  a  common  factor  of  20  a  and  25  a  ; 
2>a^i/  is  a  common  factor  of  \2o[^if'  and  Ibx^i^. 

143.  Two  expressions  which  have  no  common  factor  ex- 
cept 1,  are  said  to  be  privie  to  each  other. 

144.  The  Highest  Common  Pactor  of  two  or  more  expres- 
sions is  the  product  of  all  the  factors  common  to  the 
expressions. 

Thus,  3  a?  is  the  highest  common  factor  of  3  a^,  6  a^,  and 
I2a\ 

boi^'if  is  the  highest  common  factor  of  10a;^y^  and  Ibs^if. 

For  brevity,  H.  0.  F.  will  be  used  for  Highest  Common 
Factor. 

(1)  Find  the  H.  C.  F.  of  '^2cc'h''x  and  21a2^V. 

^2a%\x  =2xSx7Xa^xb^Xx; 
21aWx'  =  S  x7Xa^xb^Xx^. 

.-.  the  B..G.F.  =  Sx7xa^xb^Xx. 
=  2\aWx. 

(2)  Find  the  H.  C.  F.  of  2 o?x  ^2ax?  and  3 ahxy  f  3 hx?y. 

2  all-  +  2  ax?  =  2  ax  {a  +  x) ; 
3  ahxy  +  3  hx^y  =  3  hxy  (a  +  x). 

.\  the  E..C.F.^x(a  +  x). 


COMMON    FACTORS    AND    MULTIPLES.  89 

(3)    Find  the  H.  C.  F.  of 

8  aV  -  24  a^^  +  16  a^  and  12  aj?y  - 12  axy  -  24  ay, 

8aV  -  24  A  +  16^2  =  8^2  (:r2  _  3:r  +  2), 
=  2V(:r-l)(:c-2); 
12a^y  -  12aa;y  -  24ay  =  12ay  {x'-x -  2), 

=  22x3ay(:r  +  l)(a;-2). 

.-.  theH.C.F.--=22a(:r-2), 
-4a  (a; -2). 

Hence,  to  find  the  H.  C.  F.  of  two  or  more  expressions : 
Resolve  each  expression  into  its  lowest  factors. 
Select  from  these  the  lowest  power  of  each  common  factor  ^ 
and  find  the  product  of  these  poivers. 

"^^r  Exercise  XLVII. 

Find  the  H.  0.  F.  of: 
1.    18aZ>Vr/and36a2^(?r^2         g.    l?;)^^^  34/y,  and  51//. 

3.  ^:^y'z\  12:ry2«,  and  20^;^^. 

4.  30. ry,  90a:V,  and  120 ^r^^^ 

5.  a^  _  IP-  and  a«  -  h\  7.    a^  +  x^  and  {a  +  x)\ 

6.  a^  _  ^.2  ^j^(j  (^^  _  _^)2^  g^    '^o^-\QXi^{Zx-\-\f. 

9.  7:1:2  _  4^.  ^j^^  7A-4a2. 

10.  12  A^y  -  4  a3^y2  ^^^^j  39  ^2^^  _  ^q  r^Vy^. 

11.  8  a^h^'c  -  12  a%c^  and  6  a^»^c  +  4  a5V. 

12.  .i;2_2:r-3  anda,-2  +  a:-12. 

13.  2a3-2a^2^nd4^(a  +  ^))2. 

14.  12:i-3y  (a;  -y)(:^;  -  3y)  and  \^3?{x-y){^x-~y). 
16.  3a:3^e^_24-^  and  6:^3-962:. 


90  ALGEBRA. 


^. 


16.  ac  {a  —  b){a  —  c)  and  be  (b  —  a){b~  c). 

17.  I0a?y-   mot^f-i-bxfo.ndiba^f-bxf-lOOy'^. 

18.  X  {x  +  if,  x^(a^-l)  and  2x{3^-x  -  2). 

19.  35.^-6^  +  3,  6a;2_|_g^_;L2,  and  12a;2_  12. 

20.  6  (a  -  b)\  8  (a^  _  ^2)2,  and  10  (a*  -  5^). 

21.  x^-f,{x-{-yf,2.nd.x'-\-2>xy-^2f. 

22.  ^i-^  — ?/^,  a::^  — ?/^,  and  a.-^— 7a:y4-62/^. 

23.  x^-\,a^--l,Q.\idiX^  +  x-2. 


145.  When  it  is  required  to  find  the  H.  C.  F.  of  two  or 
niore  expressions  which  cannot  readily  be  resolved  into 
their  factors,  the  method  to  be  employed  is  similar  to  that  of 
the  corresponding  case  in  arithmetic.  And  as  that  method 
consists  in  obtaining  pairs  of  continually  decreasing  numbers 
which  contain  as  a  factor  the  H.  C.  F.  recpired ;  so  in  alge- 
bra, pairs  of  expressions  of  continually  decreasing  degrees 
are  obtained,  which  contain  as  a  factor  the  H.  C.  F.  re- 
quired. 

The  method  depends  upon  two  principles : 

1.  Any  factor  of  an  expression  is  a  factor  also  of  any 
multiple  of  that  expression. 

Thus,  if  ^represent  a  factor  of  an  expression  A,  so  that  A  =  nF, 
then  mA  =  mnF.     That  is,  mA  contains  the  factor  F. 

2.  Any  common  factar  of  two  expressions  is  a  factor-  of 
the  sum  or  difference  of  any  r)iuUiples  of  the  expresdons. 

Thus,  if  i^  represent  a  common  factor  of  the  expressions  A  and  B 
so  that 

A  =  mF,  and  B  =  nF; 
then  j)-^  =  pmF,  and  qB  =  qnF. 

Hence,        pA  ±  qB  =  pmF ±  qnF, 
=  (pm  ±  qn)  F. 
That  is,  pA  ±  qB  contains  the  factor  F. . 


COMMON    FACTORS    AND    MULTIPLES.  91 

146.    The   general  proof  of  this   method  as  applied  to 
numbers  is  as  follows  : 

Let  a  and  b  be  two  numbers,  of  which  a  is  the  greater. 
The  operation  may  be  represented  by  : 


b)a(p 

42)154(3 

nF)mF{p 

pb       . 

126 

pnF 

T)  b  (q 

28)42(1 

cF)  nFiq 

qc 

28 

qcF 

d)c(r 

14)28(2 

F)cF(c 

rd 

28 

cF 

p,  q,  and  r  represent  the  several  quotients, 
c  and  d  represent  the  remainders, 
and  d  is  supposed  to  be  contained  exactly  in  c. 
The  numbers  represented  are  all  integral. 
Then  c  -^  rd, 

b-=qc  +  d  =  qrd  -\-  d  =  {qr  +  \)  d, 
a-^ph  +  c  =pqrd  +  pd  +  rd, 
=  {pqr  +  p  +r)  d. 

.'.  d  is  a  common  factor  of  a  and  b. 

It  remains  to  show  that  d  is  the  highest  common  factor  of  a  and  b. 

Let/  represent  the  highest  common  factor  of  a  and  b. 

Now  c  =  a  —pb,  and/  is  a  common  factor  of  a  and  b. 

.•.  by  (2)/ is  a  factor  of  c. 

Also,  d=b  —  qc,  and  /  is  a  common  factor  of  b  and  c. 

.".  by  (2) /is  a  factor  of  d. 

That  is,  d  contains  the  highest  common  factor  of  a  and  b. 

But  it  has  been  shown  that  rf  is  a  common  factor  of  a  and  b. 

.'.  d  is  the  highest  common  factor  of  a  and  b. 

Note.  The  second  operation  represents  the  application  of  the 
method  to  a  particular  case.  The  third  operation  is  intended  to  rep- 
resent clearly  that  every  remainder  in  the  course  of  the  operation 
contains  as  a  factor  the  H.C.  F.  sought,  and  that  this  is  the  highest 
factor  common  to  that  remainder  and  the  preceding  divisor. 


92  ALGEBRA. 


147.  By  the  same  method,  find  the  H.  C.  F.  of 

2a;2  +  a;-3)4:r^  +  8a^-    x-6{2x  +  S 
4:a^-]-2x^-6x 

ex'  +  bx-G 
6a^-i-Sx-9 

2xi-^)2x'-\-    x-S{x-l 
2ar'  +  3x 
-2x-S 
.-.  the  H.C.F.  =  2a;  +  3.  -2a;-3 

The  given  expressions  are  arranged  according  to  the  descending 
powers  of  .T. 

The  expression  whose  first  term  is  of  the  lower  degree  is  taken 
for  the  divisor ;  and  each  division  is  continued  until  the  first  term  of 
the  remainder  is  of  lower  degree  than  that  of  the  divisor. 

148.  This  method  is  of  use  only  to  determine  the  com- 
pound factor  of  the  H.  C.  F.  Simple  factors  of  the  given 
expressions  must  first  be  separated  from  them,  and  the 
highest  common  factor  of  these  must  be  reserved  to  be 
multiplied  into  the  compound  factor  obtained. 

Find  the  H.  C.  F.  of 

12^;^  +  30 a,-3 -  72:^2  and  32.^«  +  84^.^ -  176a;. 

l2x'+S0x^~72x''--=6x^(23^  +  5x-12). 
32  a:^  ^  84  ^si  -i76x-^4:x(Sx^-j-  21  x  -  44). 
6x^  and  4:X  have  2x  comvion. 

2a,-2  +  5:t'- 12)8.^2  +  21  a:-44  (4 
8.1-2  + 2037-48 

X+   ^)2x''-{-bx-l2{2x-^ 
2x^  +  8x 
-3.1'- 12 
.-.  the  H.  C.  F.  =  2.1-  {x  +  4).  -3a;-12 


COMMON    FACTORS    AND    MULTIPLES.  93 

149.    Modifications  of  this  method  are  sometimes  needed. 

(1)    'FmdiheR.G.'F.oi4.ar'-8x--ba.ndl2x'-4:x-65. 

4:a^-8x-b)12x'-   4a;-65(3 
120.-^- 24  a: -15 
20  a; -50 

The  first  division  ends  here,  for  20a;  is  of  lower  degree  than  4.r'. 
But  if  20  a;  —  50  be  made  the  divisor,  4a^  will  not  contain  20  a;  an  in- 
tegral number  of  times. 

Now,  it  is  to  be  remembered  that  the  H.  C.  F.  sought  is  contained 
in  the  remainder  20  a;  —  50,  and  that  it  is  a  compound  factor.  Hence 
if  the  simple  factor  10  be  removed,  the  H.  C.  F.  must  still  be  con- 
tained in  2  a;  — 5,  and  therefore  the  process  may  be  continued  with 
2a;  —  5  for  a  divisor. 

2x-&)4:x'-    8a:-5(2a;+l 

4^^10^ 

2a;-5 
2a:-5 

.-.the  H.C.F.  =  2:r-5. 
(2)    Find  the  H.  C.  F.  of 

2lx^ ^-  4:x^ -  I5x - 2 8ind2la^ -  S22^ ~  b4:x -  7. 
21a;«-4ar^-15a;-2)21:r3-32a;2_^4^_rj.^j 

21a^-   4:x'-~15x-2 
-28a.'2-39a7-5 

The  difficulty  here  cannot  be  obviated  by  removing  a  simple  factor 
from  the  remainder,  for  -  28.^2  _  39a;  -  5  has  no  simple  factor.  In 
this  case,  the  expression  21  ar*  -  4  a-^  -  15  a;  -  2  must  be  multiplied  by 
the  simple  factor  4  to  make  its  first  term  divisible  by  —  28a<^, 

The  introduction  of  such  a  factor  can  in  no  way  aflFect  the  H.  C.  F, 
sought ;  for  the  H.  C.  F.  contains  only  factors  common  to  the  remain- 
der and  the  last  divisor^  and  4  is  not  a  factor  of  the  remainder. 

The  signs  of  all  the  terms  of  the  remainder  may  be  changed ;  for 
if  an  expression  A  is  divisible  by  —  F,  it  is  divisible  by  +  F. 


94 


ALGEBRA. 


The  process  then  is  continued  by  changing  the  signs  of  the  remain- 
der and  multiplying  the  divisor  by  4. 

28x'  +  ^9x-i-b)84:x'-    IQar"-   60x-   8(3x 

-ISSa^-    1bx-~   8 
Multiply  by  -  4,  —  4 

532^^  +  300  a:  +  32(l9 
532:^^+741  a: +95 
Divide  by  -  6S,  -  63)- 441  a;- 63 

7x-{-    1 

7a7+l)28:r^+39a:  +  5(4x  +  5 
28a^+   4:x 

35x  +  5 
.'.  tIieH.C.F.  =  7ar+l.      35.T  +  5 

<3)    Find  the  H.  C.  F.  of 

8a:^  + 2ar- 3  and  6a:*+ 5^  -  2. 
6^+    5a;2_    2 


i^+2x--3)24:a 

;'«  +  20:r^-    8   (3.r  +  7 

24a;H    &^-    9x 

Ux'^   9x~   8 

Multiply  by  4, 

4 

56:^^  +  36:^-32 

56a^+14:r-21 

Divide  by  11, 

ll)22ar-ll 

2x-    l)8a^+2x-S{4:X-{-S 

8ar'-4:x 

6x-3 

/.  the  H.  C.  F. 

=  2:r-l.                                 6:r-3 

In  this  case  it  is  necessary  to  multiply  by  4  the  (p,ven  expression 
Ga^  +  5a^  — 2  to  make  its  first  term  divisible  by  Sr*,  4  being  obvi- 
ously not  a  common  factor. 


COMMON    FACTORS    AND    MULTIPLES. 


95 


The  following  arrangement  of  the  work  will  be  found 
most  convenient : 


8:r2  +  2.r 


6x~S 
6^7-3 


6c(^+    bar" 
4 


24:0^ +  20x'-    8 
24:^+    ear"-    9x 

Ux'i-    9x~    8 

_4 

56^2  + 36a; -32 

56a;^+14a;-21 

n)  22  a; -11 

2x-    1 


3a? 

+  7 

4:X-\-S 


150.  From  the  foregoing  examples  it  will  be  seen  that,  in 
the  algebraic  process  of  finding  the  highest  common  factor, 
the  following  steps,  in  the  order  here  given,  must  be 
carefully  observed : 

I.  Simple  factors  of  the  given  expressions  are  to  be  re- 
moved from  them,  and  the  highest  common  factor  of  these 
is  to  be  reserved  as  a  factor  of  the  H.  C.  F.  sought. 

II.  The  resulting  compound  expressions  are  to  be  ar- 
ranged according  to  the  descending  powers  of  a  common 
letter ;  and  that  expression  which  is  of  the  lower  degree  is 
to  be  taken  for  the  divisor  >  or,  if  both  are  of  the  same 
degree,  that  whose  first  term  has  the  smaller  coefficient. 

III.  Each  division  is  to  be  continued  until  the  remainder 
is  of  lower  degree  than  the  divisor, 

IV.  If  the  final  remainder  of  any  division  is  found  to 
contain  a  factor  that  is  not  a  common  factor  of  the  given 
expressions,  this  factor  is  to  he  removed;  and  the  resulting 
expression  is  to  be  used  as  the  next  divisor. 

V.  A  dividend  whose  first  term  is  not  exactly  divisible 
by  the  first  term  of  the  divisor,  is  to  be  multiplied  by  such 
an  expression  as  will  make  it  thus  divisible. 


96  ALGEBRA. 


r:ih 


Exercise  XL VIII. 
Find  the  H.  C.  F.  of: 
1.    h:x?^^x-\,    %)3?-\-'21x-h. 

3.  6a*  +  25a3_21a2  +  4a,    "l^a^-^Wlo? ~^^a^ ■\-\^a. 

4.  9:1^  +  9:^:2  _  4^  _  4^    45a;3_^  54^_  20^:- 24. 

5.  272:«-3a;*  +  6a;3-3a;2^    162:c«  +  48ri^- 18:i;2_|_  g^ 

6.  20a;»- 60:^:2  _|_  50^  _  20,    a2a:^- 92:^^  + 68a:2_24a;. 
^^.  4:r2-8a;-5,    12a:2_4^_65 

8.  Za^-^c?x-'2.ax^,    9  a^  -  8  a^a:  -  20  a^r^. 

9.  '  10:^3 +  372 -9a: +  24,    20a;^- 17a:2  +  48a;  -  3. 

10.  8a:3-4ar^- 3207 -182,    360.-3 -84a;2_m^_;L26. 

11.  5a;2(;i2o73+4a72+17o;-3),  10x(24o;3_52^+14^_l), 

12.  -  9a;V  -  ^y"  -  20o:yS    ISo^^y  -  1807^/  -2xif-  d>y\ 

13.  657^-07-15,    9o:2_3^_20. 

14.  12o7»-9or2  +  5o7  +  2,    24o:2_^i0o7  +  l. 
-15.6073  +  15072-607  +  9,    9o7«  +  6o72-51o7  +  36.       V 

^16.   4o;3-o^y-o7y2_5/,    7o73  +  4o72y  +  4o7y2_  3y3.^ 

17.  2a3-2a2-3a-2,    ?>a^  -  a" -2a-ie>.  ^ 

18.  123/3  _^  23/2  _94y_  60,    483/3_  24^2 -3482/ +  30. 

19.  9o7(2o7*-6o^-o,-2+15o7-10), 

6o^(4o7^  +  6o73-4o^- 1507-15). 

20.  'l5o;*  +  2y-75o72+5o7+2,   35o7*+o73-175o^+30o7  + 1. 

21.  '21o7*-4o73- 1507^-207,    210.-3-32072-5407-7. 

22.  9o7V-22o723/3_3^^4_|.iOy5,   ^ x^y ~^ xY^o(^f~2b xf. 


COMMON    FACTORS    AND   MULTIPLES.  97 

23.    Qx'-Ax^-lla^-Sx'-dx-l,  T 

'24.   x*-aa^-  aV -  a^x  -  2a^f  3r^  -  laa^  +  3a^x  -2a^ 

151.  The  H.  C.  F.  of  three  expressions  will  be  obtained 
by  finding  the  H.  C.  F.  of  two  of  them,  and  then  of  that  and 
the  third  expression. 

For,  if  ^,  £,  and  Care  three  expressions, 

and  D  the  highest  common  factor  of  ^  and  B, 

and  U  the  highest  common  factor  of  D  and  C, 

Then  D  contains  every  factor  common  to  A  and  B, 

and  U  contains  every  factor  common  to  I)  and  C. 
.'.  E  contains  every  factor  common  to  A,  -B,  and  C. 

Exercise  XLIX. 
FindtheH.O.F.  of: 

1.  2a^  +  x-l,    cc^-^f)x  +  A,    o^'+l. 

2.  l^-y'-y  +  l,    33/2-2y-l,    f-y^-^^y-l, 

3.  2:^-4:^2+90;- 10,  c(^+2x'-^x-{-20,  3i^-^bx'-^x-^2>b. 

4.  ^-7:r2  +  16:r-12,     Zx"" -1^3^ +l(Jx, 

bx^-l^x'+lx-U. 

5.  3/«-5y2+lly-15,    f-,f  +  3y-^b, 

rt  22/^-7/+16y-15. 

6.  2x^  +  ^x-5,    Ss^-x-2,    2x^  +  x-3. 

7.  a^-1,    c(^-~a^-x-2,    2x^-x'-x-S. 

8.  57^-3^-2,    20^  +  3x^-1,    or^-l-l. 

9.  12(x*-y*),    10(x^~y^),    8(x'2/ +  X7/'). 

10.  x*-{-x7f,    x^y-j-i/,    X* -j- x-if -{- y*. 

11.  2{xhj-xy^),    ^{3?y-^xi/),    ^{x\j-xy^),    bix^y-xy"). 


98  ALGEBRA. 


Lowest  Common  Multiple. 

152.  A  common  multiple  of  two  or  more  expressions  is 
an  expression  which  is  exactly  divisible  by  each  of  them. 

153.  The  Lowest  Common  Multiple  of  two  or  more  ex- 
pressions is  the  product  of  all  the  factors  of  the  expressions, 
each  factor  being  written  with  its  highest  exponent. 

154.  The  lowest  common  multiple  of  two  expressions 
which  have  no  common  factor  will  be  their  product. 

For  brevity  L.  C.  M.  will  be  used  for  Lowest  Common 
Multiple. 

(1)  Find  the  L.  C.  M.  of  12  a\  Ubc^,  SQah\ 

145^=2  X  Ibc^, 

.-.  the  L.  C.  M.  =2'  X  32  X  Ta^^V  =  252a'b'c'. 

(2)  Find  the  L.  C.  M.  of 

2a^  +  2ax,6a^-(jx^,^a^-6ax-\-Sa^. 

2a^  +  2ax  =2a(a  +  x), 

6a2_6a:2  =  2  X  3  («  +  ^)  (a-a;), 

Sa""  -  6ax  +  dct"  =  S  (a  -  xf. 

.'.  the  L.  C.  M.  =  6  a  (a  +  x)  (a  -  xf. 

Exercise  L. 
Find  the  L.  C.  M.  of: 

1.  A.a^x,    6aV,    2ax^.  4.   a^—1,    c(?--x. 

2.  V^ax",    12af,    I2xij.  5.    a?-b\    a^-\-ab. 

3.  x",    ax  +  x".  6.    2:?;-l,    4::^-!. 


COMMON    FACTORS   AND    MULTIPLES.  99 

7.  a  +  h,    a^-{-b\  "9.    x'-x,   :^-l,   a^-\-l. 

8.  x^-l,   ^2_|_i^    ^4_i  10^    ^_2^    ^_^^   ^_1. 

11.  2a +1,    4^2- 1,    8a3+l. 

12.  {a  +  bf,    o?-b\ 

13.  4(l  +  a:),    4(l-a:),    ^il-x"). 

14.  :r-l,    a^  +  a;  +  l,    a;3_i^ 

15.  x'-f,    {x  +  y)\    {x-y)\ 

16.  x^-y^,     3(2:-y)^     12(a;«  +  y^).        .  "    ,    ,. 

17.  6(:r2+^),   8(a;y-y2),    lOC^^-y^^^  -C   ^  '^ 

18.  a.'^  +  Sa^  +  e,    a;2_|_6^_^3_ 

19.  a2_^_20,    a2  +  a-12. 

20.  :^  +  lla;  +  30,    ;r2+12a;  +  35. 

21.  o?-^x-22,    x'-Ux  +  22. 

22.  4a5(a2-3a5  +  2^>2),    ba^a'  +  ab  -  6b^. 

23.  20(a;2_i)^    24(:r2-2;-2),    16(a;^  +  a;- 2). 

24.  12:^2/ (:i;2_^)^    2:r2(a:  +  y)^     32/2(a;~y)^ 

25.  {a-b)(b-c),    (b-c)(c-a),    (c-a)(a-b). 

26.  (ct  — 5)(a  — c),    (b  —  a){b~c),    (c~a)(c  —  b). 

27.  :^-4:X^  +  3x,    x*  +  c(^-12x^,    x^  +  Sx^-ia^. 

28.  x^i/-xf,    Sx(x-7/y,    ^7j{x-yf. 

29.  {a+bf-{c-^d)\   (a^cf-{b  +  df,   {a+df-{b  +  cf. 

30.  (2a;-4)(3a;-6),   (a; -3) (4 a: -8),   (2a;-6)(5:r- 10). 

155.  "When  the  expressions  cannot  be  readily  resolved 
into  their  factors,  the  expressions  may  be  resolved  by  find- 
ing their  H.  C.  F. 


100  ALGEBRA. 


I.    Find  the  L.  0.  M.  of 


^o(^y  +  4:Xif-\-2f 
^^y  +  ^xf  +  2f 


^o(?-22xf-    Sf 
2 


18C^-UX2/-U7/ 


lly)33:r^.y-44a:y^>-22?/ 


3:r^  —  ^xy  —  2y'^2x~y 

Hence,  6^^-11  :r2y  + 2^/3  =  (2a;- y)  {?>a^-^xy-22f), 
and  9a;3_22x/-8^  =  (3a;  +  4y)(3:r2-4:ry-2y2). 

.-.  the  L.  C.  M.  =  (2:r  -  y)  (3:r  +  4y)  (3^-2  -  4:ry  -  2y2). 

In  this  example  we  find  the  H.  C.  F.  of  the  given  expres- 
sion, and  divide  each  of  them  by  the  H.  C.  F. 

156.  It  will  be  observed  that  the  product  of  the  H.  C.  F. 
and  the  L.  0.  M.  of  two  expressions  is  equal  to  the  product 
of  the  given  expressions.     For, 

Let  A  and  B  denote  the  two  expressions,  and  D  their 
H.  C.  F. 

Suppose  A  =  aD,  and  B  =  hD  ; 

Since  D  consists  of  all  the  factors  common  to  A  and  B, 
a  and  h  have  no  common  factor. 

.".  L.  C.  M.  of  a  and  h  is  ah. 

Hence,  the  L.  C.  M.  of  aD  and  hD  is  ahD. 

Now,  A  =  aD,  and  B  =  hD\ 

:.  AB  =  ahDx  D. 

AB 

:.  ——-  =  ahD  =  the  lowest  common  multiple.     That  is. 

The  L.  C.  M.  of  two  expressions  can  he  found  hy  dividing 
their  product  by  their  H.  C.  F. 

Or,  hy  dividing  one  of  the  expressions  hy  the  H.  C.  F.,  and 
multiplying  the  result  hy  the  other  expression. 


COMMON    FACTORS    AND    MULTIPLES.  101 

157.  To  find  the  L.  C.  M.  of  three  expressions,  A,  B,  0. 
Find  M,  the  L.  C.  M.  of  ^  and  ^  ;  then  the  L.  C.  M.  of  M 
and  (7  is  the  L.  C.  M.  required. 

Exercise  LI. 
Find  the  L.  CM.  of: 
1.    Q^-x-2,    21x^-l1x-\-2,    14a;2+5a;-l. 
■2.    .r^-l,    a^-{-2x-2>,    ^a^-x-2. 

3.  0^-21,    3(?-lbx  +  ?>^,    ar^-3:z;2-2a;+6. 

4.  5;r2+19^-4,    lOx'-^-l^x-?,. 

5.  I2x'  +  xy-^f,    ld>a^+\Sxy-20f. 

6.  x^-2a^  +  x,    2a:*-2a;^-2a;-2. 

7.  I2a^-^2x-^,    12^ -42a: -24,    12ri;2_  23^;- 24. 
a   x''-^x'-\-llx-Q,    a.-3-9:r2  +  26:r-24, 

9.  .t2_4^2^  a;3+2a^  +  4a2a:  +  8a^  ^-2(2a:2_^4a2a;- ^a^. 

10.  a:3_^2a;2^-^2/^-22/3^    y? ~2:k?7J ~ xf -\-2f. 

11.  1+;^+/,    l-p+lA    l+/+i?'. 

12.  (1-a),    (1-a)',    (l-«)'. 

13.  {a  4-  6^)2  -  ^>^    (a  +  ^)'  -  c",    (^  +  cf  -  a\ 

14.  3c3-3c2y  +  c2/2_^^    4c3_^y_3cy. 

15.  m^-Sm  +  S,    m«  +  3m^  +  m  +  3. 
16. '20 w^ +  71^-1,    2bn^+bn^-n-l. 

17.  ^>^-253  +  52-86  +  8,    453-12524-9^-1. 

18.  2?-^-87-*+129^-8r2  +  2r,    3r'-6r3  +  3r. 


# 


j^ 


CHAPTER    VIII. 

Fractions. 

158.  The  expression  ^  is  employed  to  indicate  that  a 

b 

units  are  divided  into  h  equal  parts,  and  that  one  of  these 
parts  is  taken ; 

or,  that  one  unit  is  divided  into  h  equal  parts,  and  that  a 
of  these  parts  are  taken. 

159.  The  expression  ~  is  called  a  fraction,     a  is  the  nu- 

h 

merator,  and  h  the  denominator. 

160.  The  numerator  and  denominator  are  called  the 
terms  of  the  fraction. 

161.  The  denominator  shows  into  how  many  equal  parts 
the  unit  is -divided,  and  therefore  names  the  part ;  and  the 
numerator  shows  how  many  of  these  parts  are  taken. 

It  will  be  observed  that  a  letter  written  above  the  line 
in  a  fraction  serves  a  very  different  purpose  from  that  of  a 
letter  written  behw  the  line. 

A  letter  written  above  the  line  denotes  number ; 

A  letter  written  below  the  line  denotes  name. 

162.  Every  whole  number  may  be  written  in  the  form 
of  a  fraction  with  unity  for  its  denominator  ;  thus,  a  =  -. 


FRACTIONS.  103 


To  Eeduce  a  Fraction  to  its  Lowest  Terms. 

163.  Let  the  line  AB  be  divided  into  5  equal  parts,  at 
the  points  C,  D,  U,  F. 

A I    I    I    I    I    I    I    I    I    I    I    I    I    I    I    It? 
C  D  E  F 

1\Q^AF\s.\oiAB.  -  (1) 

Now  let  each  of  the  parts  be  subdivided  into  3  equal 
parts. 

Then  AB  contains  15  of  these  subdivisions,  and  ^i^  con- 
tains 12  of  these  subdivisions. 

.-.  ^i^isifof  ^^.  •  (2) 

Comparing  (1)  and  (2),  it  is  evident  that  |-  =  xl"- 

In  general: 

If  we  suppose  ^^  to  be  divided  into  h  equal  parts,  and 
that  ^i^  contains  a  of  these  parts, 

Then  J^i^is^of^^.  (3) 

Now,  if  we  suppose  each  of  the  parts  to  be  subdivided 
into  G  equal  parts. 

Then  AB  contains  he  of  these  subdivisions,  and  AF  con- 
tains ac  of  these  subdivisions. 

.-.  ^i^is^of^^.  (4) 

he 

Comparing  (3)  and  (4),  it  is  evident  that 

a oc 

h~  he 

Since  —  is  obtained  by  multiplying  by  c  both  terms  of 

the  fraction  -, 
h 

and,  conversely,  -  is  obtained  by  dividing  by  e  both  terms 

of  the  fraction  ~,  it  follows  that 
he 


104  ALGEBRA. 

I.  If  the  numerator  and  denominator  of  a  fraction  be 
multiplied  by  the  same  number,  the  value  of  the  fraction 
is  not  altered. 

II.  If  the  numerator  and  denominator  be  divided  by  the 
same  number,  the  value  of  the  fraction  is  not  altered. 

Hence,  to  reduce  a  fraction  to  lower  terms, 

Divide  the  numercdor  and  denominator  hy  any  common 
factor. 


164.  A  fraction  is  expressed  in  its  lowest  terms  when 
both  numerator  and  denominator  are  divided  by  their 
H.  0.  F. 

Reduce  the  following  fractions  to  their  lowest  terms  : 
(1) 


o?  —  o?  _  (a  —  x)  (a^  -\-ax-^2^)  __  a^-\-ax-{-a^ 
o?  —  a?  {a  —  x){a-\-x)  a-\-x 


(2) 


(3) 


(4) 


a^+7a+10  ^  (a  +  5)(«  +  2)  ^  a  +  5 
a2  +  5a  +  6        (a  +  3)(a  +  2)       a  +  3" 

^x'-bx-^  _  (2a; -3) (3 a; +  2)  ^  3ic-f  2 
8^-2a:-15       (2a; -3)  (4a; +  5)       4a:  +  5" 

a^-7a^+16^-12 
3a3-14a2+16a  * 


Since  in  Ex.  (4)  no  common  factor  can  be  determined 
by  inspection,  it  is  necessary  to  find  the  H.  C.  F.  of  the 
numerator  and  denominator  by  the  method  of  division. 

Suppress  the  factor  a  of  the  denominator  and  proceed  to  divide : 


FRACTIONS. 


105 


a^-    7a^+16a~    12 


3 

3a« 
3a« 

-21a2  +  48a- 
-14a2+16a 

36 

3 

36 

-21a2  +  96a^ 
-21a2  +  98a- 

108 
112 

-2j-    2a  + 

4 

3a2 
3a^ 


14a  +  16 
6a 


~    8a+16 
-   Sa  +  16 


a-7 
3a  — 8 


.-.  tlieH.C.F.  =  a-2. 

Now,  if  a^  -~7a^-\-16a  —  12  be  divided  hj  a-2,  the  re- 
sult is  a^  —  5 a  +  6 ;  and  if  3a^  — 14:0,^  +  16a  be  divided  by 
a  —  2,  the  result  is  3  a^  —  8  a. 

.   a^-7a^+16a-12^a^-5a  +  6 
3a«-14a2  4-i6a  3a2_8a  ' 

165.  When  common  factors  cannot  be  determined  by  in- 
spection, the  H.  C.  F.  must  be  found  by  the  method  of 
division. 

Exercise  LII. 

Reduce  to  lowest  terms  : 


1.    - 


3. 


5. 


4:X(X'i-l) 

c^-9x  +  20 
^'2_7a;+12* 

:z.'2-10:c+21* 

x^  +  x'+l 

C(^  +  X-\-l' 


6. 


a^+l 


a^  +  2a'  +  2a+l 


a  — 20 


a^  +  a-12 


'  a^  +  2x''-3x  +  20' 


9. 


10. 


a;^-53;^+ll^-15 
x^-a^  +  Sx  +  b    ' 

x^  -f  o?y  -\-  x}^  —  y* 
X*  —  o^y  —  xi/  —  y^' 


r^:^-^ 


106  ALGEBRA. 


a;3  +  a;2_^_;^  ^2_^^_2^,2 

8a;3-27a^       "  '       o?  +  ab-ac 

^g    15a^  +  a5-25^  6:g^-lla:^y  +  3^y' 


a^  +  2  a5  +  52  -  c2  (a  -  5)^  -  (c  +  df 

x*  —  a^  —  2x  +  2  6a;^  — 5a:  — 6 

*       2a;3_^_l     •  •  8a;2_2^_i5' 

a;^-6a;^+ll^-6  oc      ^*  +  ^/  +  l/* 

'     x^-2a^-x  +  2   '      ^  '  (x-y)(a^-f)' 

^g    6a^-23x'+16x-3  ^         x^  +  y^ 


6:r3-17:i,-2+lla:-2  x'~ary'  +  7/ 

x'-a^-x  +  l  (a'  +  b')(a'  +  ah-{-b') 

'  x''~23?-x^~2x-\-\  '  (a^-b')(a'-ab  +  b'y 

To  Reduce  a  Fraction  to  an  Integral  or  Mixed 
Expression. 

x^4- 1 
Change  — ^^— -  to  a  mixed  expression. 
X  —  1 

(a^+l-)^(x-l)  =  x'^  +  x+l  +  -^.     Hence, 

X  —  1 

166.  If  the  degree  of  the  numerator  of  o.  fraction  eqybals  or 
exceeds  that  of  the  denominator^  the  fraction  m.ay  be  changed 
to  the  form  of  a  nfiixed  or  integral  expression  by  dividing  the 
numerator  by  the  denominator. 


FRACTIONS.  107 


The  quotient  will  be  tlie  integral  expression,  the  remainder 
(if  any)  will  be  the  numerator,  and  the  divisor  the  denom- 
inator, of  the  fractional  expression. 

Exercise  LIII. 
Change  to  integral  or  mixed  expressions : 

,    x'~2x-i-l  ^    lOaJ'-llax  +  lOx^ 

1. .  o.  — 


X 


—  1  ba  —  x 


z,  ', •  *•   — ', ^i — • 

x-\-^  4a:  — 1 

^    ^■a^-\-e>x-\-b  o    2:^-bx-2 

d. .  o. . 

a;  +  4  x  —  4: 

^    a^-ax  +  x'  g    ^_±^ 

a-\-  X  a  —  b 

2^_±_5  y.     b:^-x'-\-b 

'     x-3'  bx'-\-4:X-l' 


To  Eeduce  a  Mixed  Expression  to   the  Form  of  a 
Fraction. 

167.    In  arithmetic  5|  means  5  4-1. 

But  in  algebra  the  fraction  connected  with  the  integral 
expression,  as  well  as  the  integral  expression,  may  be  posi- 
tive or  negative ;  so  that  a  mixed  expression  may  occur  in 
any  one  of  the  following  forms : 


,  a  a  ,  a  a 


108  ALGEBRA. 


Change  ?2  -|-  -  to  a  fractional  form. 

Since  there  are  b  bths  in  1,  in  n  there  will  be  n  times  b 
bths,  that  is,  nb  bths,  which,  with  the  additional  a  bths, 
make  nb-{-a  bths. 


■••"U-   h  ■ 

In  like  manner : 

a      nb~a 

""  b      b   • 

- 

-+f==^"^ 

and  - 

a      —  nb  —  a 

""  -b~    b    ■ 

Hence, 


168.  To  reduce  a  mixed  expression  to  a  fractional  form, 
Multiply  the  integral  expression  by  the  denominator,  to 

the  product  annex   the  numerator,  and  under  the  resuU 
write  the  denominator. 

169.  It  will  be  seen  that  the  sign  before  the  fraction  is 
transferred  to  the  numerator  when  the  mixed  expression  is 
reduced  to  the  fractional  form,  for  the  denominator  shows 
only  what  part  of  the  numerator  is  to  be  added  or 
subtracted. 

The  dividing  line  has  the  force  of  a  vinculum  or  paren- 
thesis affecting  the  numerator ;  therefore  if  a  minus  sign 
precede  the  dividing  line,  and  this  line  be  removed,  the 
dgn  of  every  term  of  the  numerator  must  be  changed. 
Thus, 

a  —  b      cn  —  {a~b)      en  —  a  +  6 


FRACTIONS.  109 


X—1 

(1)    Change  to  fractional  form  x-~l-\ . 


X  — 

■^^~ 

1 

a?- 

-x  +  {x- 

-1) 

X 

^- 

~x-\-x- 

-1 

x 

^- 

-1 

x~\ 

(2)    Change  to  fractional  form  x  —  \ 


x- 

X 

a?~ 

-x-{x-l) 

X 

x"- 

-x-x+l 

X 



-2x-\-l 

Exercise  LIV. 
Change  to  fractional  form  : 

2.    1+^-^ 


3.    Zx 


x-\-y 

X 

a^-\-x' 


u. 

UUU  Zj 

^       5a- 

■66' 

6. 

a-\-h- 

a^  +  52 
a-\-h' 

7. 

7        2- 

-3a  +  4, 

a« 

5-6a 

* 

a 

o        5. 
c>x 

0^  —  3 

4.   a  —  a;  + 

a  — a;  ^  2a 


110  ALGEBRA. 


9.    ±tb  +  l.  15.    2a~b~^. 

a  —  o  a-\-o 


10.  ^-1.  1&   3^-10  ■     41 


a-\-b  37  +  4 

11.   -^^-(x  +  y).  17.   :^  +  ^+l+     ^ 


x-j-i/  X—  1 

4  a:-2 

13.  a-l  +  -J— .  19.    a^  — 2aa;-f  4a;2 ?— . 

a+1  a-|-2a; 

14.  ^  +  5-2^.  20.    ^_a  +  2/  +  ^^^T=^- 


Lowest  Common  Denominator. 

170.  To  reduce  fractions  to  equivalent  fractions  having 
the  lowest  common  denominator : 

3  5;    2  ?y  5 

Reduce  — -,  -^,  and  - — -   to   equivalent   fractions   hav- 
4a^    2>a  Q>a^ 

ino;  the  lowest  common  denominator. 

The  L.  C.  M.  of  4a^  3  a,  and  6a«  =  12  a^ 

I  f  both  terms  of  —4  be  multiplied  by  3  a,  the  value  of 
the  fraction  will  not  be  altered,  but  the  form  will  be 
changed  to ^  ;  if  both  terms  of  -^  be  multiplied  by  4  a^, 

the  equivalent  fraction  — — ^  is  obtained  ;  and,  if  both  terms 
12  a^ 

5  •  .        10 

of  — ^  be  multiplied  by  2,  the  equivalent  fraction  — — ^  is 

obtained. 


FRACTIONS.  Ill 


XT                                ^x     2?/       5 
Hence,  — -,    — ^,    

are  equal  to  ^3,    |g,    jl^,  respectively. 

The  multipliers  3  a,  4a^,  and  2,  are  obtained  by  dividing 
12  a^  the  L.  C.  M.  of  the  denominators,  by  the  respective 
denominators  of  the  given  fractions. 

171.  Therefore,  to  reduce  fractions  to  equivalent  frac- 
tions having  the  lowest  common  denominator, 

Find  the  Ij.Q.VL.  of  the  denominators. 

Divide  theJj.O.yi.  hy  the  denominator  of  each  fraction. 

Multiply  the  first  numerator  hy  the  first  quotient,  the  sec- 
ond hy  the  second  quotient,  and  so  on. 

The  products  will  he  the  numerators  of  the  equivalent 
fractions. 

The  L.  C.  M.  of  the  given  denoTninators  will  he  the  derwrn- 
inator  of  each  of  the  equivalent  fractions. 

Exercise  LV. 

Eeduce  to  equivalent  fractions  with  the  lowest  common 
denominator : 

,     ^x-1   ^x-9  ^  1  1 

1.    ::: ,  — tt: — •  5. 


6     ' 

18 

2 

x--4cy 

?>x-^y 

bx"     ' 

'      10:1' 

4 

a  —  bc 

3a-2t? 

bac 

I2a^c  ■ 

5 

6 

{a  —  h){h  —  c)  {a  —  h){a  —  c) 

g         ^a?  xy 

'   3  (a +  5)'  6(a2-^2y 


3.       „        .    :"    r  .         7. 


8 07  +  2  2a;-l    Zx-]-2 
x-2  '  3^7  —  6'  5a;  — 10' 


.        ^  Kj  -    a  —  hm   -,    c  —  hn 


;'  l-x"' 


7nx  nx 


112  ALGEBRA. 


Addition  and  Subtraction  of  Fractions. 

172.  To  add  fractions: 

Reduce  the  fractions  to  equivalent  fractions  having  the 
lowest  common  denominator. 

Add  the  numerators  of  the  equivalent  fractions. 
Write  the  result  over  the  lowest  common  denominator. 

173.  To  subtract  one  fraction  from  another : 

Reduce  the  fractions  to  equivalent  fractions  having  the 
lowest  common  deno7)iinator. 

Subtract  the  number  ator  of  the  subtrahend  from  the  numer- 
ator of  the  Tninuend. 

Write  the  result  over  the  lowest  common  denom,inator. 

(1)  Simplify,       i2-!lI+3A=li. 

The  lowest  common  denominator  (L.  C.  D.)  =  15. 
The  multipUers  are  3  and  1  respectively. 

12  a;  +  21  =  1st  numerator, 
3  re —    4  ^=  2d  numerator, 

\bx  -{-VJ  =  sum  of  numerators. 

.    4^+7   ■   3:y  — 4^  15a;4-17 
"       5  15  15      ' 

.ON    o-      vr  3a  — 4Z>       2a-~b-\-c  ,  13(2  —  4(7 

(2)  Simplify,       —^ 3""+ 12 * 

The  L.  CD.  =84. 

The  multipliera  are  12,  28,  and  7  respectively. 

36  a  —  48  5  =  1st  numerator, 

—56  a.+  28  ^  —  28  c  =  2d  numerator, 

91  a  —  28<?  =  3d  numerator. 

71a  —  20 6  —  56(?  =  sum  of  numerators. 

•    3a-45       2a~b  +  c  .    13a- 4c  _71a- 20^>  -  56g 
"         1  3  "^        12  84 


FRACTIONS.  113 


Since  the  mmus  sign  precedes  the  second  fraction,  the  signs  of  all 
the  terms  of  the  numerator  of  this  fraction  are  changed  after  being 
multiplied  by  28.       '    ' 


Exercise  LVI. 
Simplify : 

3a;  — 2y       5x  —  1i/      Sx-\-2y 
'   ~Tx        ^      10^      "^       25 


4^-7y^       3^-8y       5-2y 
3^       "^       6a:       "^       12     * 


^aJ'+bb'      Sa-{-2b      7-2a 
2P       '^       bb       ~^      9     ' 


4ar-f5  _  Sx-1   ,    _9_ 
8  5x  Via?' 


^x~Zy      3a;+7y  _  hx-~2y       9x  +  2j/ 
'         7        "^       14  21       "^       42     * 


g    3 0:3/ -4  __  5y^+7  _  6a;^-ll 
ar'y^  a;?/^  a^^y 

a^-2ag  +  g^       Z/^-2Z>g  +  g» 


5a^-2       3a^-a 
•       8^2  8      * 


Q    a  —  ^,^  —  t?,tf  —  a,    a^^  -{-b(?-\  ca^ 

y. 1 1 7 — I . 

cab  aoc 


10    J: I J_      g^r^z      y-2z 

■  2^"^      6j^z      2:1-3^        42^s*        4a^yz' 


114  ALGEBRA. 

Simplify  ^~^  4-  ^±^. 
x^y      x—y 

The  L.  CD.  =^-3/2. 

The  multiphers  are  x  —  y  and  a;  +  y  respectively. 

a?  —  2  a;y  +    3/^  =  1st  numerator, 
3?  -\-  '^xy  -\'    ?/^  =  2d  numerator. 
2a;^  +  2?/^  =  sum  of  numerators, 

or,  K^^f)  =    "     "         " 

"  a;  +  y       a;-^         ^  —  ^ 

Exercise  LVII. 
Simplify : 

1,    -^  +  -L-.  6. 

a:  —  6       a;  +  5 


x—1       x  —  o 


1 4-^       1  —  x 


4.^ ^.  9. 


1 

,            1 

2a{a  +  x) 

'    2a  {a - 

-X) 

a 

h 

{a-\-b)b 

{a  —  h)  a 

5 

3 

2x{x-l) 

4:X{X- 

2)' 

l  +  x 

l-x 

l-x       1-s^'  l+x  +  oc^       l-x  +  x"' 

1  X  Q     2ax  —  Sby  _  2ax-\-^hy 

\   x-y      {x - yf  '   2xy{x-y)       2xy(x-\-y) 

,,x    a-      v£,    2a-\-b       2a  — b         6ab 

(1)    Simplify  4 -^ ^ — 7^. 

^  a  — ^  a  +  b        c?  —  b^ 

TheL.C..D.  =  (a-Z))(a  +  ^>). 

The  multipliers  are  a  +  6,  a  —  6,  and  1,  respectively. 


FRACTIONS.  115 


2a^  -\-  S  ab  -\-  b^  ^=  1st  numerator, 

2a^  -\-  S ab  —  P  =  2d  numerator. 

—  6ab  ^  3d  numerator. 


0  =  sum  of  numerators. 

.   2a-\~b       2a  — b         Qab     _^ 
' '    a  —  b         a-\-b        o?  —  b^ 


The  L.  C.  D.  =  {x  +  y)  (x-y)  {x^  +  y^). 

The    multipliers    are    a^  +  if,     {x  —  y)  (a^  +  y*),     (a;  +  y)  (x  ~  y) 
(a^  +  y^),    {x  +y){x  —  y),  respectively. 


+  y^  =  1st  numerator, 

~x^-\-2  o^y  —  2  7?if-  +  2  X'l^  —  y^  =  2d  numerator, 

X*  —  y*  =  3d  numerator, 

2c(^y  — 2x1^  =  4tli  numerator. 

4  xi^y  —     ct^y^  —  2/^  =  ^^^  ^^  numerators. 


.'.  Sum  of  fractions 


_  ia^y  —  a^y^  —  y'^ 


Simplify 


x*  —  y^ 
Exercise  LVIII. 


^        1       ■       1       ■      2a  g    __^ ^!— 4-      ^ 


l-\-a       1  —  a       1—a^'  l~x       1  —  x       l-^af'' 

1  1       ,      2:r  .    X   ,       y       ,        x^ 


1—x     l-\-x      1  +  ^  y     ^  +  y     ^  +  ^y 

f,   X  —  1  r  ^  —  ^  t  ^  —  ^ 

a;-2  ~^  x~^  "^  :i:  — 4* 

.        3       ,       4a  5a2 

6. f- 


X  —  a       (^  —  «)^       (^  —  ciy 


116  ALGEBRA. 


7.        ^  ^ 


x-l      x  +  2       {x  +  l)(x+2) 
a  —  b  . b  —  c , 


(b  +  c)(c  +  a)       (c-\-a)(a  +  b)       (a+b)ib  +  c) 

X  —  a       x~b  (a  — by 

y.   Y  H~ 


X—  b       x  —  a       (x  —  a){x-~b) 
10.   ^  +  y  _    ^^    _|.    x^y-:^ 

y       x+y     y{p^-f) 
Jj 


{b  —  c)  {c  —  a)       {c  —  a)  [a  —  b)       {a  —b){b  —  c) 

^„  G?  —  bc  ,  b^  —  ac  ,  c^  +  a5 

•L&'    ', — : — TTT — : — ^    \    Ti — : — rr; — r^r   \ 


13. 


14. 


{a-\-b){a  +  c)       {b  +  a){b-\-c)       {c -\- b)  {c -\- a) 

a  X  c?  -\-  o(? 

a  —  x       a-{-2x       (a  — x)  (a-\-2x)' 

3 4  ,  6 

{a-b){b-c)       {a-b){a-c)'^  {a-c){b-c)' 

x  —  2y  2x-\-y  2x 


a  —  b     __     a  —  b  {a~b){x-^y) 

'   x{a-\-b)       y(a-{-b)  xy{a-\-b) 

17.      3^         ^  +  2y  ,       3y 

a  — c  a  — 5 


{a  +  bf-c"       {a  +  cf-b^' 

19.    ^+^  _  ^  —  ^    I  ^^(^— y) 

aa;  +  ^y       ax— by       c^3?  —  5y 


FRACTIONS.  117 


174.    Since  ~  =■  «,  and  — ^  =  a, 

b  —0 

it  is  evident  that  if  the  signs  of  both  numerator  and  de- 
nominator be  changed,  the  value    of  the  fraction  is  not 

altered. 

a  —  h       —{a—-h)       —a-\-b_b—a 


Again, 


c  —  d      -~{c  —  d)       —c-\-d      d—o 


Therefore,  if  the  numerator  or  denominator  be  a  com- 
pound expression,  or  if  both  be  compound  expressions,  the 
sign  of  every  term  in  the  denominator  may  be  changed, 
provided  the  sign  of  every  term  in  the  numerator  be  also 
changed. 

Since  the  change  of  the  sign  before  the  fraction  is  equiva- 
lent to  the  change  of  the  sign  before  every  term  of  the 
numerator  of  the  fraction,  the  sign  before  every  term  of 
the  denominator  may  be  changed^  provided  the  sign  before 
the  fraction  be  changed. 

Since,  also,  the  product  of  +  «  multiplied  by  +  5  is  ab, 
and  the  product  of  —  a  multiplied  by  —  5  is  ab,  the  signs 
of  two  factors,  or  of  any  even  number  of  factors,  of  the  de- 
nominator of  a  fraction  may  be  changed  without  altering 
the  value  of  the  fraction. 

By  the  application  of  these  principles,  fractions  may  often 
be  changed  to  a  more  simple  form  for  addition  or  subtrac- 
tion. 

(1)    Simplify  2  _       3^^  |£-3 
X      2x—l      1— 4:r 

Change  the  signs  before  the  terms  of  the  denominator  of  the  third 
fraction,  and  change  the  sign  before  the  fraction. 
The  result  is, 

2_       3  2x--S 

X      2x-l       ^a^-V 

in  which  the  several  denominators  are  written  in  symmetrical  form. 


118 


ALGEBRA. 


The  L.  C.  D.  =  ^  (2:^:  -  1)  (2 :?:  +  1). 

8c(^  —  2     =  1st  numerator, 

—  6:r^  —  3:r  =  2d  numerator, 

—  2ar^-}-3x  =  M  numerator. 


.*.  Sum  of  tlie  fractions  = 


—  2     =^  sum  of  numerators, 
-2 


a;(2a;-l)(2a;  +  iy 
(2)    Simplify 

1        +,.,    I,    .+        ' 


a  {a  —  h)  {a  —  c)       b  (b  ~  a)  {h~c)       c{c  —  a)  (c  —  b)' 

Change  the  sign  of  the  factor  {b  —  a)  in  the  denominator  of  the 
second  fraction,  and  change  the  sign  before  the  fraction. 

Then  change  the  signs  of  the  factors  (c  —  a)  and  (c  —  h)  in  the  de- 
nominator of  the  third  fraction. 

The  result  is, 

1  1^1 


a(a  —  b)(a  —  c)       b{a  —  b)(b  —  c)       c(a  —  c){b-~cy 

in  which  the  factors  of  the  several  denominators  are  written  in  syra 
metrical  form. 

The  L.  C.  D.  =  abc  {a  -b)(a-  c)  (b  -  c). 

be  (b  —  c)  =  b^c  —  bc^  =  1st  numerator, 

—  ac  (a  —  c)  =  ~  a^c  -f-  (^c^  =  2d  numerator, 

ab  (a  —  b)  =  a^b  —  aP  =  3d  numerator. 

a^b  —  a^c  — •  ab^  +  ac^  -\-  b^o  —  bc^  =^  sum  of  numerators. 
=  a^{b-c)~a  (b^  -  c^)  +bc(b-  c), 
=  [a"  ~a(b  +  c)  +  be]  [b  ~  c], 
=  [a^  —  ab  —  ac  -p  ^c]  [b  —  c], 
=  [(a^  -  ac)  ^  (ab  -  be)]  [b  -  e], 
=  [a(a  —  c)  —  b(a  —  c)]  [b  —  c], 
=  (a  —  b)(a  —  e)  (b  —  e). 


FRACTIONS.  119 


.*.  Sum  of  the  fractions  =  -^^ ^y^- ^-L — L^ 

aoc  {a  —  0){a  —  c)  {b  —  c) 

abc 

Exercise  LIX. 
Simplify : 

1.    ^   I  ^~y 

^-y     y  — ^' 

3  +  2a;   .    Sx-2   ,    16a;  — :g^ 
2-a;  "^   2  +  x  "^    .^^-4  * 


o;^-!       ^  +  1       1-x 


3-3y2      2-2y      6y  +  6" 

1  2.1 


(2-m)(3-m)       (m-l)(m-3)       (m-l)(m-2) 


(b-a)(x  +  a)       (a-b)(x  +  by 
a^-b^~^  b^-a^~^  a^  +  b^' 


g    b  —  a__a~2b       Sx(a~b) 
x  —  b         b-^x  b^~a^    ' 


3  +  2a:       2-3^;       lear-a:^ 
2-:z;         2-\-x  ^-4  * 


10     _J 7  4-20:^' 

l~2a;       1  +  2^;        4:0^-1' 


120  ALGEBRA. 


-  -  a-\-b I b-\-c I c-{-a 

(b  —  c)(c  —  a)       (b  —  a)ia~  c)      (a  —  b){b~  c)' 

12  G^  —  bc         ,  y^  -\-aG         ,  (?  ^ab 


{a-b){a~c)       ib-\-c){b-a)       {c~a){c-\-b) 

13.       y+^       J        g+^       I       ^+y 


14. 


{x-y){x  —  z)       {y-x){y  —  z)       {z  —  x){z~-y)' 

3 4 6 

{a-~b){b-c)       {b-a){c-a)       {a-c){c-b)' 


15.    . \ + I ±. 


# 


Multiplication  of  Fractions. 


175.  Hitherto  in  fractions,  equal  parts  of  one  or  more 
units  have  been  taken.  But  it  is  often  necessary  to  take 
equal  parts  oi  fractions  of  units. 

Suppose  it  is  required  to  take  f  of  -I  of  a  unit. 

Let  the  line  AB  represent  the  unit  of  length. 

>il    I    1    I    I    I    I    I    I    I    I    I    I    I    I    1^ 
0  D  E  F 

Suppose  AB  divided  into  5  equal  parts,  at  C,  D,  E, 
and  F,  and  each  of  these  parts  to  be  subdivided  into  3 
equal  subdivisions. 

Then  one  of  the  parts,  as  AC,  will  contain  3  of  these  sub- 
divisions, and  the  whole  line  AB  will  contain  15  of  these 
subdivisions. 

That  is,  -|-  of  -|-  of  the  line  will  be  -^  of  the  line  ; 

I  of  I-  will  be  J^  +  ^i^-f- J,  + J^,  or  -5^,  of  the  line;    and 

I  of  I  will  be  twice  3^,  or  ^,  of  the  line. 


FRACTIONS.  121 


Suppose  it  is  required  to  take  -  of  -  of  the  line  AB. 

d       0 

>tl    I    I    I    I    I    I    I    I    I    I    I    I    I    I    Iff 
O  D  E  F 

Let  the  line  AB  be  divided  into  h  equal  parts,  and  let 

each  of  these  parts  be  subdivided  into  d  equal  subdivisions. 

Then  the  whole  line  will  contain  hd  of  these  subdivisions, 

and  one  of  these  subdivisions  will  be  — -  of  the  line. 

bd 

If  one  of  the  subdivisions  be  taken  from  each  of  a  parts, 

fchej  will  together  be  —-  of  the  line.    That  is, 
bd 

~  of  —  = [ [ taken  a  times,  =  — , 

d       b       bd     bd     bd  bd 

and  -  of  -  will  be  c  times  — -,  or  -—  of  the  line. 
d      b  bd        bd 

Therefore,  to  find  a  fraction  of  a  fraction, 

Find  the  product  of  the  numerators  for  the  numerator  of 
the  product,  and  of  the  denominators  for  the  denominator 
of  the  product. 

116,    Now,  -  X  -  means  -  of  -. 
d      b  do 

Therefore,  to  find  the  product  of  two  fractions, 

Mnd  the  product  of  the  numerators  for  the  numerator  of 

the  product,  and  of  the  denominators  for  the  denominator 

of  the  product. 

The  same  rule  will  hold  when  more  than  two  fractions 
are  taken. 

If  a  factor  exist  in  both  a  numerator  and  a  denominator, 
it  may  be  cancelled  ;  for  the  cancelling  of  a  common  factoB 
before  the  multiplication  is  evidently  equivalent  to  cancel- 
ling it  after  the  multiplication ;  and  this  may  be  done  by 
§163. 


122  ALGEBRA. 


Division  op  Fractions. 

177.  Multiplying  by  the  reciprocal  of  a  number  is  equiv- 
alent to  dividing  by  the  number.  Thus,  multiplying  by  J 
is  equivalent  to  dividing  by  4. 

The  reciprocal  of  a  fraction  is  the  fraction  with  its  terms 
interchanged. 

Thus,  the  reciprocal  of  |  is  |,  for  |  X  |  =  1.  §  42. 

Therefore,  to  divide  by  a  fraction. 

Interchange  the  terms  of  the  fraction  and  multiply  by  the 
resulting  fraction.     Thus : 

(V\    2a    .     1   ^  2a       3a?  ^  2a 
The  common  factor  cancelled  is  3  a?. 

^  ^    277/2  •   9y      272/2      7^      g^- 
The  common  factors  cancelled  are  9y  and  7x. 

(v\         g3?      _^  _  ab ax  (a  +  ^K^~  ^3 

{a~xy      to"  — x^       {a  —  x){a~x)  ah 

x{a-\-x) 
b  {a  —  x)' 
The  common  factors  cancelled  are  a  and  a  —  x. 

If  the  divisor  be  an  integral  expression,  it  may  be  changed 
to  the  fractional  form.  §  162.- 

EXEECISE   LX. 

1.    A  X  — .  3.       ^^      --    ^^ 

'   bx       d'  2p  ~2  '  p  —  l' 

2^       3a^      3a£  8:rV  _^  ^ 

'     a  ^     c         26'  *   Ibab^  '  Sab^' 


FRACTIONS.  123 


'   45:r2y  ^  2^a^b^'  8//        2a:y       90  w^' 

9:r^y^2        _20^  25^       TOti^^       3p7i 

lOa^^^c  ^       18  ^ry^^'  *    147iV       75/m      4/lV 

^^   3^  ^  5y^  ^  _  12^  ^^      a~b    ^  a'~h\ 

4:x^       6x7/            2x7/^  a^-i-ab       a/^  —  ab 

a'  +  b'  _  a^  a^  +  x-2      a;^-13a;  +  42 

•  a?-b^  '  a  +  b'  x^-7x             x'  +  2x. 

j^Tx'-llx  +  SO      a^-Sx  a^-a^       (a  +  xY 

x'-Qx  +  d        a^-5x  a'  +  a^      (a-xf 

15.    2^(^^-/y  ^  ^        ■ 

ex  {x-y){x  +  yf 

a'-\-2ab       ab-2b'  ^g    x'  +  xy       {x-yf 

•  a2  +  462        a^-W  '     x-y         x^-y^' 

„     c(?  —  A:        ar^  — 25  .    m^~n^  _  n  —  m 


r^  +  5a;      :r^  +  2a;  (^  +  d^        c-\-d 

20    ^'-^«  +  3        a'-9a  +  20        a^~la 
'   a2-5a  +  4       a2_i0a  +  21       a^-Sa" 

Z>^-7&  +  6      Z>^+10^  +  24  ^  ^>^  +  6^> 
•   5^  +  35-4      5^-146  +  48  '   53-86^* 

22.      ^-y"     X  ^y~'^y^ X  ^~^^ 

r^-3^y  +  2y2        ar'  +  ary        ipo—yf 
a^~Sa^b  +  SaP-b^  .   2a5-2Z>^      a^  +  a5 


124  ALGEBRA. 


24. 


^    {x-af-l^a^-{b-ay 


26. 


(x-hf-a?      x^~{a-hf 

{a-\-hy-{c  +  df   .   {a-cf-{d-hf 

\2  ~ 


{a  +  cf-ih  +  ay  '   {a-bf-id-cf 


27.   ^-2^^  +  3/^-2',,  ^  +  3/-g 

Complex  Fractions.  \ 

178.  A  complex  fraction  is  one  which  has  a  fraction  in 
the  numerator  or  in  the  denominator,  or  in  both. 

179.  A  fraction  may  be  regarded  as  the  quotient  of  the 
numerator  divided  by  the  denominator. 

This  is  the  simplest  meaning  of  a  complex  fraction. 
Therefore,  to  simplify  a  complex  fraction, 
Divide  the  numerator  by  the  denominator. 

(1)    Simplify  4. 
2 


f  =  i^i  =  ix|=|. 

(2) 

Simplify  ||. 

3  =  |  =  |-^¥  =  |xA-=i%. 

(3) 

Simplify     ^^   . 
x  —  \ 

?>X      _        ^X                ?,X     ,     4:X~1          Zx  ,,          4 

x-\      ^x-l        1*4            1   '^  4.^-1 
4      _    12. 

4a:-l 


FRACTIONS.  125 


It  is  often  shorter  to  multiply  both  terms  of  the  fraction 
by  the  L.  C.  D.  of  the  fractions  contained  in  the  numerator 
and  denominator. 

Thus,  in  (1),  multiply  both  terms  by  6  ;  in  (2),  both  terms 
by  24 ;  in  (3),  by  4.       The   results   obtained   are   |,    j^-^, 

respectively. 


4:X~l 

(4)   Simplify 


l+x 


l-x  +  x" 


x(l—x^a^) 


1    1^1  ^  (l-{-x){l~x  +  x')-i-x 

^    ^l-x  +  x" 


X 


1      x  —  x^-j-a^ 
1  +  x  +  a^ 


a;  (1  +  ^  +  ^) 


l-i-x  +  s^  — (x  — a^-\-a^) 

x-{-ct^-{-x* 


The    expression    is    reduced    to    the 

form  xg-x  +  a^) ^^.^^  ^  x-^  +  a^^ 

{l+x)(l-~x  +  x')-^x  l+a;  +  :r3 

The  expression is  reduced  to  the  form 

1      x  —  x^-^-xr 

l  +  x  +  oi? 


126  ALGEBRA. 

Exercise  LXI. 
Simplify : 

Sx  ,   x—1 

1.  _2_I_3__  s_   1 1_ 

^(^+l)-|-2i  1  +  1 


X 


:r—  1 


2 1^  9.    1  + 


07-2  +  -^  1+^-f--^ 


:r  —  6 
3.    -^ 2£-l  ,j, 


^+1         ^_L?_1  1 L 


:r^  + 


2     2  1  +  1 


11. 1 


^_(x-b){x-c} 


x  +  a  ^_^^^    2^ 


1- 


\a;      aj  \x  '  aj                 ,o    V^      ^        J  \x      a        J 
lA ~ 

/a      xy 

\x      a) 


-J       X  —  a 

x-\-a 

1  X 


x-y      o^-ff  -x^+-T~W         ^2+  r 


a,-^— 2/^      x-]-y\{x—yY      x—y  J 


xy-\-f      x^  +  xy 


x  —  y 


£+1      ;r  - 1  {^-f){^i:?-2xy) 

a;+l_a;  — 1        '  '  -t?/ 


FRACTIONS.  127 


ab ac 

0^ -\- {a -\- h)  X -{- ah      x^ -\- {a -\- c)  x -\- ac 


b-C 


x'+{b  +  c)x-\-hc 

a~{-h  ■       h 

16.  -^+1 L_  17.  -i 2±i 

1+1       -+i  1+1 

X  a      0 

,     2m-3  +  i  ^J-  +  l  +  2- 

18.    ^       19.    ^^      ^"      ^"       20.   L_ 

2m-l  g^  -  (5  +  g)' 


1  + 


m  ab  14^ 


#= 


1- 

EXERCISE    LXII. 
MISCELLANEOUS   EXAMPLES. 


2.  Find  the  value  of  ^' +  f  ~  "'' +  ^^^  when  a  =  4,b==i, 

1  o?  —  W  —  e  -\-2bc 

3.  Find  the  value  of  3a^  -j -^  when  a  =  4,  5  =  i, 

c  =  l.  "^         ^ 

4.  Simplify        2  11 


(a;2-l)2      2.2r2-4a7  +  2      l-a;2- 

X  X 

6.  Find  the  value  of  {^-=4\~ ^~^^  +  ^  when  :.  =  ^±^ 

\x  —  b)      x-\-a^2b  2    * 

7.  Simplify  j-^±l---^-^    4--2JLl^^ 


128 


ALGEBRA. 


9.    Simplify 

yU-y     y^V^/'     A^+^y+2/^     y 


-1 


/3     y\^  +  ^y+y^ 

10.    Simplify 

/a^-a5\/a^  +  a^>  +  ^'\  ,  /   2a^    -  - 1  Vl  -  - 


2aZ> 


+  a6  +  Z^' 


11.    Simplify 


a-{-x 


1  + 


a;^ 


1- 


c?  —  a? 


a-\-x  €?-\-^ 

12.   Divide  ^  +  i-3(^i-^)  +  4(:.  +  ^)by^+l 
1  22:2/ 


13.    Simplify 


1-f 


1  + 


X} 


14,  Find  the  value  of  f±^  +  ^^--|^  when  ^ 

,  'Ih  —  x      2b4-x     4:b^  —  ar 

_    ab 

~  a  +  b' 

15.  Find  the  value  of  ^"^^~^    when   a;  =  ^,"^  ,     and 


y 


ab-}-a 


ab  +  1 
16.    Simplify 


+ 


x  —  y  +  l 


1 


a5  +  l 


+ 


a{a  —  l)(a  —  c)       b{b  —  c)  (b  —  a)       c{c  —  a)  (c  —  b) 

a-1      b-1      c-1 
^abc  a  b  c 


YI,    Simplify 


be  -\-ca  —  ab 


1+1-1 

a      b      c 


FRACTIONS.  129 


'^  2  2 

n  i7r  —  IT 


18.   Simplify    — '1 X 


1       1        "  m^  +  n^ 
n     m 


\  ' 


19.  Simplify    |_±t£|l  +  ^±i!z:An 

a      b-\-c 

20.  Simplify    3a-[6  +  S2a-(5-.)J]  +  |  +  |^ 

_J: L_4_ ^ ^2/ 


2L    Simplify    a-rr      g-y      (a-^r)^      (g-y^ 
ia-y){a-xf      {a-x){a-yf 

22.    Simplify    I 23.    (-' ~l/')(!^^ -^-V) 

1,^+1  ^^      x-\-y 


24. 


Simplify    (<l^-<t^^(<l±l^<t±^. 


25.  Simplify     l -f. '^  + ^+y 

{x-y){x-z)      {y~x){y~z)      {z-x){z-y) 

26.  Simplify     \ -I-  ^ L 

a{a~h){a~-c)      h(b  —  a){b  —  c)      abc 

27.  Simplify    2±ix  ^~^ 


. 6_         (:r-l)(^-2) 

^-1 


9?^ 


4P 

y  CHAPTER  IX. 

Feactional  Equations. 
to  reduce  equations  containing  fractions. 

180.   (1)  f +-f  =  12. 

Multiply  both  sides  by  4,  the  L.  C.  M.  of  the  denominators. 
Then,  2  a;  +  a;  =  48, 

3a;  =  48, 
.-.  a;  =  16. 

(2)   |-4=24-| 


Multipl; 

y  both  sides 

by 

24, 

the  L.  C.  M.  of  the  denominators. 

Then, 

4a; -96  =  576- 3a;, 
4x4- 3a;  =  576 +96, 
7  a;  =  672, 
.-.  X  =  96. 

(3)| 

x-1 
11 

X  - 

-9, 

' 

Multiply  by  33,  the  L.  C.  M.  of  the  denominators. 

Then,  11a;- 3a; -t- 3  =  33a; -297, 

11a;  _  3^  _  33a;  =  -297 -3, 
-25a;  =  -300, 
.-.  x  =  12. 

Since  the  minus  sign  precedes  the  second  fraction,  in  removing' 
the  denominator,  the  +  (understood)  before  x,  the  first  term  of  the 
numerator,  is  changed  to  — ,  and  the  —  before  1,  the  second  term  of 
the  numerator,  is  changed  to  +. 

181.    Therefore,  to  clear  an  equation  of  fractions, 
Multiply  each  term  hy  the  L.  C.  M.  of  the  denominators. 


FRACTIONAL  EQUATIONS.  131 

If  a  fraction  is  preceded  by  a  minus  sign,  the  sign  of 
every  term  of  the  numerator  must  he  changed  when  the 
dcTiominator  is  removed. 

Exercise  LXIII. 

Solve  the  equations : 

1    5^     ^  +  ^^71  4.   5a:      5a;^9      2>  —  x 

2  ■  2        4       4         2' 

3  3  6  5 

3. 


5-2a?  ■  o^^      6a;-8   ^    3:  +  2_14      3  +  5a; 


2  2  9  4 

-    5ar4-3       3-4a;   .   a;  ^  31       9-5a; 

8  3       "^2       2  6     • 

8.  15^_6-z7^^o(.-l). 

9.  ^^-2^+7=3:.- 14. 


10. 


2  3 

7a;  +  5      5a;-6^8-5a; 
6  4  12     • 


11  a?  +  4      g  — 4_^g  ■  3a:  — 1 
•       3  5  "^     15     • 

^^  3^+5_2^+7_^-^Q_3^^Q^ 

13.  i(3a:-4)  +  i(5:r  +  3)  =  43-5a;. 

14.  l(27-2^)  =  |-i(7^-54). 


; 


132  ALGEBRA. 


15.   5x-\Sci;-S[16-Qx-(4:-5x)]l=6. 

507-3      9-x_5x  ■  19 

7  3  2  "^  6 

2  37  +  7      9^-8._^-ll 


16.    5a;-3_9-^^5g      19.    _^. 
7  3  2        6^         ^ 


"■        7  11  2     • 

„    8:r-15      ll:r  — l_7a7  +  2 

lo. 


19. 


3  7  13 

7a;  +  9      3^+l_9^-13      249 


^ 


7  4  14 


182.  If  the  denominators  contain  both  simple  and  com- 
pound expressions,  it  is  best  to  remove  the  simple  expressions 
first,  and  then  each  compound  expression  in  turn.  After 
each  multiplication  the  result  should  be  reduced  to  the 
simplest  form. 

8;r  +  5  ■  7^7-3  _4a;  +  6 

^^       14     "^^6:^  +  2  7      ' 

Multiply  both  sides  by  14. 

Then,  8a;  +  5  +  ^^^~^^  =Sx-\- 12. 

3x  +  l 

49  a;  — 21 

Transpose  and  combine, =  7. 

^  Sx  +  l 

Multiply  by  3a; +  1,         49 a;- 21  =  21a;  +  7, 

28  a;  =  28, 

.-.  a;  =  l. 

3_l£  ^-3 

(2)  i  =  l_i 

^^      4  4         10 

Simplify  the  complex  fractions  by  multiplying  both  terms  of  each 

fraction  by  9. 

T,,  27 -4a;      1      7a; -27 

^^'"'  -^6-^4 90— 

Multiply  both  sides  by  180. 

135 -20a;  =  45 -14a; +  54, 
_6a;  =  -36, 


FEACTIONAL  EQUATIONS.  133 


^^       Exercise  LXIV. 
Solve  the  equations : 

36  bx-4:"^^ 


2. 


9(2:r-3)  .   lla;-l_9a:+ll 
14  3:z:+l  7      ' 

103:  +  17       l2x-\-2  _bx~4: 
18  13a;- 16  9      ' 

6a;+13       3a;  +  5  _2a; 
15  5a;-25       5* 

„    18a;-22  ,  o     ,l  +  16a:_.5       101-G4a; 
'•    39^:67  +  ^'^  +  ~2r~-^^  24""- 

g    ^-bx        n-2x^   __l  +  3a;      10a;-ll  .     1 
15         14  (a; -1)  21  30  105* 

^    9a;  +  5  .  8a;- 7  _^  36a;  +  15      41 
14     "^6a;  +  2  56       "^56' 

g    6a;4-7      2a;-2_2a7+l 
'       15  7a;-6  5      * 

^    6a;+l       2a;-4       2a;--l 


10. 


15 

7a;  -  16           5     ' 

7a;-6 

a;  — 5          X 

35 

6a;- 101      5* 

183.  Literal  equations  are  equations  in  wHcli  all  the 
numbers  are  represented  by  letters ;  the  numbers  regarded 
as  known  numbers  are  usually  represented  by  the  fir^t  let- 
ters of  the  alphabet.  -?fe^ 


134  ALGEBRA. 


(1)    (a  —  x){a-^x)  =  'lo?^'2.ax  —  3?. 

Then,    a? -:^  =  2€?  ^2ax-^, 
—  2ax  =  a^, 

a 

.  .  X  = . 

2 

C2)    (x  —  a){x  —  h)  —  {x  —  h) (x  —  c)  =  2{x  —  a) {a  —  c). 

{3?  —  ax  —  hx  +  ah)  —  {3^  —  bx  —  ex  +  bc)  =  2  (ax  —  cx—a^  +  ac), 
a^  —  ax  —  bx  +  ab  —  x^  +  bx  +  cx  —  bc  =  2ax—2cx  —  2a^  +  2ac. 
That  is,  —^ax  +  3cx  =  —  2a^  +  2ac  —  ab  +  bc, 
—  2>{a  —  c)x^—2a{a  —  c)  —  b{a  —  c), 
-Zx=-2a-b, 
.    .     2a+5 

Exercise  LXV. 
Solve  the  equations : 
1,    ax-\-hc==hx-{-ac.  2.    2a ~ ex  =  Be— 5 hx. 

3.  oj^x  -\-bx  —  c  =  Px  -\-  ex  —  d. 

4.  —  a(?  +  h^e  +  ahex  ^=  aba  -{-  C7nx  —  ac^x  -\-  b^e  —  tyic. 

5.  {a-\-  X  ■\-b^{cb-\-b  —  x^  ■=  {a  -\-  x)(J)  —  x)  —  ab. 

6.  {a?-\-x)^  =  a?-\-^o?-\-a\ 

7.  {a?-x){(:i?  +  x)  =  a^-\-2ax~o^, 

8.  ^^^^  +  ^^eLdl^.  10.    ax-^-^-^=\.   . 

c  <?  2  2 


M 


^ 

—  a 

a~ 

X 

2x 

3 

bx 
ab- 

b 
-x"  _ 

\x 

b 
—  ac 

^       _      _      „ a 

12.    -^ 

X 


14.   am  —  b  —  -^~\ =  0. 

b       m 


FRACTIONAL  EQUATIONS.  135 

f^    Sax  —  2  b      ax  —  a ax      2 

Sb  2b     ^  b       3' 

ifi     ab-{-x      b^  -~x x  —  b      ab  —  x 

bx-\-\      aix'-V)  ^^    ab      ,     ,    ,  ,  1 

17.    ax ■ —  =  -^ -.  19.    —  =bc-\-d-\--. 

XX  X  X 

1&    _i?£L  +  „  +  !^  =  0.  20.    ^i^^=a.  +  ^. 

o  ~~cx  c  dx  a 


Exercise  LXVI. 

Solve  the  equations : 

x~S     __    x  —  b      ■   1 
4(a;-l)~6(a;-l)      9' 


^ 


1. 


x—  1  x-\-l 

3        7      ^6^+1'    3(l  +  2a;^) 
x-l       x+l  a^-l     ' 

4.  -^ L_ ^        ^-1 

2(a;-3)      3(rr-2)      (2;  -  2)  (:r  -  3)' 

g    ^      2(2^;  +  3)  „     6 bx  +  l 

'  9(1~x)       7-x     4:(7-xy 

6.   -IL_4  =  §i21_+2£)_^Q 
x  +  S  Sx-{-9 

„    x-1      2:«;-15  1 


x+1       2.r-6       2(a;+7) 
32:  +  5^    ^      2a:  +  3 


136  ALGEBRA. 


132^jfJ.   ,  8£+_5  ^  .  2       11    3a:-l      4a:-2__l 
•     3a;+l    "^  a;-l  ■         *    2x-l      Zx~2      6' 

10.    _^  +  ^_  =  _A_.     12.       3         .'  +  1_    :^ 


2x-?>      x-2      3:r+2'  .^-1      x—\      l-x" 

x  —  4:      x  —  b_x  —  1      rr  — 8 
x  —  b      x  —  ^      x  —  ^      X  —  9 

14.  {x  —  a)  {x  —  h)  =  {x  —  a  —  hf. 

15.  (a  —  b){x  —  c)~(b  —  c)(x—a)—(c—a)(x~b)  =  0, 

17.  __i_+_L_=. 3^_. 

:r  +  2      :r  +  3      a.^  +  5a;  +  6 

18.  {x+iy  =  x[Q~(l-x)]-2. 

^Q    25_-i^      16^+4i__23_      ^ 
a;+l  3:r  +  2        ^+1 

^   20     3a5^  a^Z>^         (2 a  +  ^)  ^^-^  _  3  .^.  ,  ^'^^ 

'    a  +  6      («  +  ^)'        a(a  +  6)'  a' 

21        4  3  29_       2 

*   .^^-8      2ru-16      24      3rr-24' 


xT--?]-^- 


22.    5-Wl-gVg-^"-^^-H 


2  +  ^+^ 


23.    1 


5      a; 


34.  ,,£:zl-+^4;4,=i  ■        1 


1(^-1)  '  f(.r  +  l)  15^1  _i 


t 


CHAPTER  X. 

Problems. 
Exercise    LXVII. 


Ex.  Find  the  number  the  sum  of  whose  third  and  fourth 
parts  is  equal  to  12. 

Let  X  =  the  number. 

Then  -  =  the  third  part  of  the  number, 

and  -  =  the  fourth  part  of  the  number, 

.*.    -  +  -  =  the  sura  of  the  two  parts. 
3     4  ^ 

But  .   12  =  the  sum  of  the  two  parts, 

•       E  4-  ? 

"34 
Multiply  both  sides  by  12: 

4a;  +  3a;  =144,  * 

7  a?  =144, 
.-.  a;  =  20f 

1.  Find  the  number  whose  third  and  fourth  parts  together 

make  14. 

2.  Find  the  number  whose  third  part  exceeds  its  fourth 

part  by  14. 

3.  The  half,  fourth,  and  fifth    of  a   certain  number  are 

together  equal  to  76  ;  find  the  number. 

4.  Find  the  number  whose  double  exceeds  its  half  by  12. 

5.  Divide  60  into  two  such  parts  that  a  seventh  of  one 

part  may  be  equal  to  an  eighth  of  the  other. 


138  ALGEBRA. 


6.  Divide  50  into  two  such  parts  that  a  fourth  of  one  part 

increased  by  five-sixths  of  the  other  part  may  be 
equal  to  40. 

7.  Divide  100  into  two  such  parts  that  a  fourth  of  one 

part  diminished  by  a  third  of  the  other  part  may  be 
equal  to  11. 

8.  The  sum  of  the  fourth,  fifth,  and  sixth  parts  of  a  cer- 
•    tain  number  exceeds  the  half  of  the  number  by  112. 

What  is  the  number  ? 

9.  The  sum  of  two  numbers  is  5760,  and  their  difference  is 

equal  to  one-third  of  the  greater.  What  are  the 
numbers  ? 

10.  Divide   45   into   two    such   parts   that   the   first   part 

divided  by  2  shall  be  equal  to  the  second  part  mul- 
tiplied by  2. 

11.  Find  a  number  such  that  the  sum  of  its  fifth  and  its 

seventh  parts  shall  exceed  the  difference  of  its  fourth 
and  its  seventh  parts  by  99. 

12.  In  a  mixture  of  wine  and  water,  the  wine  was  25  gal- 

lons more  than  half  of  the  mixture,  and  the  water 

5  gallons  less  than  one-third  of  the  mixture.  How 
many  gallons  were  there  of  each  ? 

13.  In   a  certain  weight   of  gunpowder  the  saltpetre  was 

6  pounds  more  than  half  of  the  weight,  the  sulphur 
5  pounds  less  than  the  third,  and  the  charcoal  3 
pounds  less  than  the  fourth  of  the  weight.  How 
many  pounds  were  there  of  each  ? 

14.  Divide  46  into  two  parts  such  that   if  one   part   be 

divided  by  7,  and  the  other  by  3,  the  sum  of  the 
quotients  shall  be  10. 


r 


I  » 


PROBLEMS.  139 


15.  A  house  and  garden  cost  $  850,  and  five  times  tlie  price 

of  the  house  was  equal  to  twelve  times  the  price  of 
the  garden.     What  is  the  price  of  each  ? 

16.  A  man  leaves  the  half  of  his  property  to  his  wife,  a  sixth 

to  each  of  his  two  children,  a  twelfth  to  his  brother, 
and  the  remainder,  amounting  to  §600,  to  his  sister. 
What  was  the  amount  of  his  property  ? 

17.  The  sum  of  two  numbers  is  a  and  their  difference  is  h ; 

find  the  numbers. 

18.  Find  two  numbers  of  which  the  sum  is  70,  such  that 

the  first  divided  by  the  second  gives  2  as  a  quotient 
and  1  as  a  remainder. 

19.  Find  two  numbers  of  which  the  difference  is  25,  such 

that  the  second  divided  by  the  first  gives  4  as  a  quo- 
tient and  4  as  a  remainder. 

^  .   . 

20.  Divide  the  number  208  into  two  parts  such  that  the 

sum  of  the  fourth  of  the  greater  and  the  third  of  the 
smaller  is  less  by  4  than  four  times  the  difference  of 
the  two  parts. 

21.  Find  four  consecutive  numbers  whose  sum  is  82. 

Note  I.  It  is  to  be  remembered  that  if  x  represent  a  person's  age 
at  the  present  time,  his  age  a  years  ago  will  be  represented  by  a;  —  a, 
and  a  years  hence  by  a;  +  a. 

Ex.  In  eight  years  a  boy  will  be  three  times  as  old  as  he 
was  eight  years  ago.     How  old  is  he  ? 
Let  X  =  the  number  of  years  of  his  age. 

Then  x  —  8  =  the  number  of  years  of  his  age  eight  years  ago, 
and         X  +  8  =  the  number  of  years  of  his  age  eight  years  hence, 
.-.  a;  +  8    =3(a;-8), 
x  +  S    ==3x-24:, 
a;-3x  =  -2I-8, 
-2a;  =  -32, 
a; -16. 


140  ALGEBRA. 


22.  A  is  72  years  old,  and  B's  age  is  two-thirds  of  A's. 

How  long  is  it  since  A  was  five  times  as  old  as  B  ? 

23.  A  mother  is  70  years  old,  her  daughter  is  half  that  age. 

How  long  is  it  since  the  mother  w^as  three  and  one- 
third  times  as  old  as  the  daughter  ? 

24.  A  father  is  three  times  as  old  as  the  son;  four  years 

ago  the  father  was   four   times   as  old  as  the  son 
then  was.     What  is  the  age  of  each  ? 

25.  A  is  twice  as  old  as  B,  and  seven  years  ago  their  united 

ages  amounted  to  as  many  years  as  now  represent 
the  age  of  A.     Find  the  ages  of  A  and  B. 

26.  The  sum  of  the  ages  of  a  father  and  son  is  half  what  it 

will  be  in  29  years ;  the  difference  is  one-third  what 
the  sum  will  be  in  20  years.    What  is  the  age  of  each  ? 

Note  II.  If  A  can  do  a  piece  of  work  in  x  days,  the  part  of  the 
work  that  he  can  do  in  one  day  will  be  represented  by  -^.  Thus,  if  he 
can  do  the  work  in  5  days,  in  1  day  he  can  do  i  of  the  work. 

Ex.  A  can  do  a  piece  of  work  in  5  days,  and  B  can  do  it 
in  4  days.  ^How  long  will  it  take  A  and  B  together 
to  do  the  work  ? 

Let       X  =  the  number  of  days  it  will  take  A  and  B  together. 
Then     |  =  the  part  they  can  do  in  one  day. 
Now,    J  =  the  part  A  can  do  in  one  day, 
and  \  =  the  part  B  can  do  in  one  day. 

•"•  i  +  i  =  ^^^  P^^t  A  and  B  can  do  in  one  day. 

•••    i  +  i  =  i. 
4a;  +  5a;  =  20, 
9a;  =  20, 
x  =  2l 

Therefore  they  will  do  the  work  in  2|-  days. 

27.  A  can  do  a  piece  of  work  in  5  days,  B  in  6  days,  and 

C  in  7  J  days ;  in  what  time  will  they  do  it,  all  work- 
ing togetherj' 


PROBLEMS.  141 


28.  A  can  do  a  piece  of  work  in  2^-  days,  B  in  3^  days,  and 

0  in  04-  days  ;  in  what  time  will  they  do  it,  all  work- 
ing together  ?  -    . 

29.  Two  men  who  can  separately  do  a  piece  of  work  in  15 

days  and  16  days,  can,  with  the  help  of  another,  do 
it  in  6  days.  How  long  would  it  take  the  third  man 
to  do  it  alone  ? 

30.  A  can  do  half  as  much  work  as  B,  B  can  do  half  as 

much  as  0,  and  together  they  can  complete  a  piece 
of  work  in  24  days.  In  what  time  can  each  alone 
complete  the  work  ? 

31.  A  does  |-  of  a  piece  of  work  in  10  days,  when  B  comes 

to  help  him,  and  they  finish  the  work  in  3  days 
more.  How  long  would  it  have  taken  B  alone  to  do 
the  whole  work? 

32.  A  and  B  together  can  reap  a  field  in  12  hours,  A  and 

C  in  16  hours,  and  A  by  himself  in  20  hours.  In 
what  time  can  B  and  0  together  reap  it?  In  what 
time  can  A,  B,  and  0  together  reap  it  ? 

33.  A  and  B  together  can  do  a  piece  of  work  in  12  days, 

A  and  C  in  15  days,  B  and  0  in  20  days.  In  what 
time  can  they  do  it,  all  working  together  ? 

Note  III.  If  a  pipe  can  fill  a  vessel  in  x  hours,  the  part  of  the 
vessel  filled  by  it  in  one  hour  will  be  represented  by  |.  Thus,  if  a 
pipe  will  fill  a  vessel  in  3  hours,  in  1  hour  it  will  fill  ^  of  the  vessel. 

34.  A  tank  can  be  filled  by  two  pipes  in  24  minutes  and  30 

minutes  respectively,  and  emptied  by  a  third  in  20 
minutes.  In  what  time  will  it  be  filled  if  all  three 
are  running  together  ? 

35.  A  tank  can  b^  filled  in  15  minutes  by  two  pipes,  A  and 

B,  running  together.     After  A  has  been  running  by 


142  ALGEBRA. 


itself  for  5  minutes,  B  is  also  turned  on,  and  the  tank 
is  filled  in  13  minutes  more.  In  what  time  may  it 
be  filled  by  each  pipe  separately  ? 

36.  A  cistern  could  be  filled  by  two  pipes  in  6  hours  and  8 

hours  respectively,  and  could  be  emptied  by  a  third 
in  12  hours.  In  what  time  would  the  cistern  be 
filled  if  the  pipes  were  all  running  together  ? 

37.  A  tank  can  be  filled  by  three  pipes  in  1  hour  and  20 

minutes,  3  hours  and  20  minutes,  and  5  hours, ^re- 
spectively. In  what  time  will  the  tank  be  filled 
when  all  three  pipes  are  running  together  ? 

38.  If  three  pipes  can  fill  a  cistern  in  a,  h,  and  c  minutes, 

respectively,  in  what  time  will  it  be  filled  by  all 
three  running  together  ? 

39.  The  capacity  of  a  cistern  is  755^  gallons.     The  cistern 

has  three  pipes,  of  which  the  first  lets  in  12  gallons 
in  3  J  minutes,  the  second  15-|-  gallons  in  2|-  minutes, 
the  third  17  gallons  in  3  minutes.  In  what  time 
will  the  cistern  be  filled  by  the  three  pipes  running 
together  ? 

Note  IV.    In  questions  involving  distance,  time,  and  rate : 

Distance      rp- 

— — =  iime. 

Rate 

Thus,  if  a  man  travels  40  miles  at  the  rate  of  4  miles  an  hour, 
— -  =  number  of  hours  required. 

Ex.  A  courier  who  goes  at  the  rate  of  31-J-  miles  in  5  hours, 
is  followed,  after  8  hours,  by  another  w^ho  goes  at  the 
rate  of  22|-  miles  irf  3  hours.  In  how  many  hours 
will  the  second  overtake  the  first  ? 

Since  the  first  goes  31^  miles  in  5  hours,  his  rate  per  hour  is  6j% 
miles. 


PROBLEMS.  143 


Since  the  second  goes  22J  miles  in  3  hours,  his  rate  per  hour  is  7J 
miles. 

Let  X  =  the  number  of  hours  the  first  is  travelling. 

Then  a;  —  8  =  the  number  of  hours  the  second  is  travelling. 
Then  6-^^  x  =  the  number  of  miles  the  first  travels  ; 
{x  —  8)  7^  =  the  number  of  miles  the  second  travels. 
They  both  travel  the  same  distance, 

.-.  Q>^\x  =  {x~^)1l. 

The  solution  of  which  gives  42  hours. 

4C.  A  sets  out  and  travels  at  the  rate  of  7  miles  in  5  hours. 
Eight  hours  afterwards,  B  sets  out  from  the  same 
place  and  travels  in  the  same  direction,  at  the  rate 
of  5  miles  in  3  hours.  In  how  many  hours  will  B 
overtake  A  ? 

41.  A  person  walks  to  the  top  of  a  mountain  at  the  rate 

of  2\  miles  an  hour,  and  down  the  same  way  at  the 
rate  of  3|-  miles  an  hour,  and  is  out  5  hours.  How 
far  is  it  to  the  top  of  the  mountain  ? 

42.  A  person  has  a  hours  at  his  disposal.     How  far  may  he 

ride  in  a  coach  which  travels  h  miles  an  hour,  so  as 
to  return  home  in  time,  walking  back  at  the  rate  of  c 
miles  an  hour  ? 

43.  The  distance  between  London  and  Edinburgh  is  360 

miles.  One  traveller  starts  from  Edinburgh  and 
travels  at  the  rate  of  10  miles  an  hour ;  another 
starts  at  the  same  time  from  London,  and  travels  at 
the  rate  of  8  miles  an  hour.  How  far  from  London 
will  they  meet  ? 

44.  Two  persons  set  out  from  the  same  place  in  opposite 

directions.  The  rate  of  one  of  them  per  hour  is  a 
mile  less  than  double  that  of  the  other,  and  in  4 
hours  they  are  32  miles  apart.  Determine  their 
rates. 


* 


144 


ALGEBEA. 


45.  In  going  a  certain  distance,  a  train  travelling  35  miles 
an  hour  takes  2  hours  less  than  one  travelling  25 
miles  an  hour.     Determine  the  distance. 

Note  V.   In  problems  relating  to  clocks,  it  is  to  be  observed  that 
the  minute-hand  moves  twelve  times  as  fast  as  the  hour-hand. 

Ex.  Find  the  time  between  two  and  three  o'clock  when  the 
hands  of  a  clock  are : 
I.    Together. 

II.   At  right  angles  to  each  other. 
III.    Opposite  to  each  other. 


Fig.  1. 


Fig.  2. 


Fig.  3. 


I.  Let  CH"and  CM  (Fig.  1)  denote  the  positions  of  the  hour  and 
minute  hands  at  2  o'clock,  and  CB  the  position  of  both  hands  vnen 
together. 

Then  arc  HB  =  one-twelfth  of  arc  MB. 

Let  X  =  number  of  minute-spaces  in  arc  MB. 

Then  —  =  number  of  minute-spaces  in  arc  HB, 

and  10  =  number  of  minute-spaces  in  arc  MS. 

Now  arc  MB  =  arc  MH  +  arc  I£B. 
That  is,         a;  =  10  +  -. 
The  solution  of  this  equation  gives  x  =  10^^. 
Hence,  the  time  is  10-^  minutes  past  2  o'clock. 

II.  Let  CB  and  CD  (Fig.  2)  denote  the  positions  of  the  hour  and 
minute  hands  when  at  right  angles  to  each  other. 


PROBLEMS.  145 


Let  X  =  number  of  minute-spaces  in  arc  MHBD. 

Then  —  =  number  of  minute-spaces  in  arc  HB, 

12 

and  10  =  number  of  minute-spaces  in  arc  MH. 

15  =  number  of  minute-spaces  in  arc  BD. 

Now  arc  MHBD  =  arcs  MH  +  HB  +  BD. 

That  is,  a;  =  10 +^+15. 

The  solution  of  this  equation  gives  x  =»  27-]^. 
Hence,  the  time  is  27^  minutes'  past  2  o'clock. 

III.    Let  CB  and  CD  (Fig.  3)  denote  the  positions  of  the  hour  and 
minute  hands  when  opposite  to  each  other. 

Let  X  =  number  of  minute-spaces  in  arc  MHBD. 

Then  —  =  number  of  minute-spaces  in  arc  HB, 

and  10  =  number  of  minute-spaces  in  arc  MH, 

30  =  number  of  minnte-spaces  in  arc  BD. 
Now  arc  MHBD  =  arcs  MH  ^  HB  -h  BD. 

That  is.  re  =10 +-^-1-30. 

12 
The  solution  of  this  equation  gives  x  =  43j^. 

Hence,  the  time  is  43^  minutes  past  2  o'clock. 

^ 

46.  At  what  time  are  the  hands  of  a  watch  together : 

I.    Between  3  and  4  ? 
II.    Between  6  and  7? 
^  III.    Between  9  and  10? 

47.  At  what  time  are  the  hands  of  a  watch  at  right  angles : 

I.    Between  3  and  4  ? 

II.    Between  4  and  5  ? 

III.    Between  7  and  8? 

48.  At  what  time  are  the  hands  of  a  watch  opposite  to 

each  other : 

I.  Between  1  and  2  ? 

II.  Between  4  and  5  ? 

III.  Between  8  and  9? 


146  ALGEBRA. 


49.  It  is  between  2  and  3  o'clock ;  but  a  person  looking  at 
his  watch  and  mistaking  the  hour-hand  for  the 
minute  hand,  fancies  that  the  time  of  day  is  55 
minutes  earlier  than  it  really  is.  What  is  the  true 
time? 

-^  Note  VI.  It  is  to  be  observed  that  if  a  represent  the  number  of 
feet  in  the  length  of  a  step  or  leap,  and  x  the  number  of  steps  or  leaps 
taken,  then  ax  will  represent  the  number  of  feet  in  the  distance 
made. 

Ex.  A  hare  takes  4  leaps  to  a  greyhound's  3 ;  but  2  of  the 
greyhound's  leaps  are  equivalent  to  3  of  the  hare's. 
The  hare  has  a  start  of  50  leaps.  How  many  leaps 
must  the  greyhound  take  to  catch  the  hare  ? 

Let  3  a;  =  the  number  of  leaps  taken  by  the  greyhound. 
Then  4  a;  =  the  number  of  leaps  of  the  hare  in  the  same  time. 
Also,  let  a  denote  the  number  of  feet  in  one  leap  of  the  hare. 

Then  —  will  denote  the  number  of  feet  in  one  leap  of  the  grey- 
hound. 


That  is, 

3.X^f^ 

=  the  whole  distance. 

and 

(50  + 4a;)  a- 
.    9aa; 
"     2 

=  the  whole  distance, 
=  (50  + 4a;)  a. 

Divide  by  a 

9a; 
'2" 

=  50  + 4a;, 

9a;  = 

x  = 

.-.  3a;  = 

=  100  + 8a;, 
=  100, 
=  300. 

Thus  the  greyhound  must  take  300  leaps. 

50.  A  hare  takes  6  leaps  to  a  dog's  5,  and  7  of  the  dog's 
leaps  are  equivalent  to  9  of  the  hare's.  The  hare 
has  a  start  of  50  of  her  own  leaps.  How  many  leaps 
will  the  hare  take  before  she  is  caught  ? 


PROBLEMS.  147 


51.  A  greyhound  makes  3  leaps  while  a  hare  makes  4 ;  but 

2  of  the  greyhound's  leaps  are  equivalent  to  3  of  the 
hare's.  The  hare  has  a  start  of  50  of  the  greyhound's 
leaps.  How  many  leaps  does  each  take  before  the 
hare  is  caught  ? 

52.  A  greyhound  makes  two  leaps  while  a  hare  makes  3 ; 

but  1  leap  of  the  greyhound  is  equivalent  to  2  of  the 
hare's.     The  hare  has  a  start  of  80  of  her  own  leaps. 
How  many  leaps  will   the  hare  take  before  she  is 
^^ caught  ?    V  t^  0 

Note  VII.  It  is  to  be  observed  that  if  the  number  of  units  in  the 
breadth  and  length  of  a  rectangle  be  represented  by  x  and  x  +  a, 
respectively,  then  x{x  -{■  a)  will  represent  the  number  of  surface  units 
in  the  rectangle,  the  unit  of  surface  having  the  same  name  as  the 
linear  unit  in  which  the  sides  of  the  rectangle  are  expressed. 

53.  A  rectangle  whose  length  is  5  feet  more  than  its  breadth 

would  have  its  area  increased  by  22  feet  if  its  length 
and  breadth  were  each  made  a  foot  more.  Find  its 
dimensions. 

54.  A  rectangle  has  its  length  and  breadth  respectively  5 

feet  longer  and  3  feet  shorter  than  the  side  of  the 
equivalent  square.     Find  its  area. 

55.  The  length  of  a  rectangle  is  an  inch  less  than  double 

its  breadth ;  and  when  a  strip  3  inches  wide  is  cut 
off  all  round,  the  area  is  diminished  by  210  inches. 
Find  the  size  of  the  rectangle  at  first. 

56.  The  length  of  a  floor  exceeds  the  breadth  by  4  feet ;  if 

each  dimension  were  increased  by  1  foot,  the  area 
of  the  room  would  be  increased  by  27  square  feet. 
Find  its  dimensions. 

Note  VIII.  It  is  to  be  observed  that  if  h  pounds  of  metal  lose  a 
pounds  when  weighed  in  water,  1  pound  will  lose  I  of  a  pounds, 
or  I  of  a  pound. 


148  ALGEBRA. 


57.  A  mass  of  tin  and  lead  weighing  180  pounds  loses  21 

.pounds  when  weighed  in  water ;  and  it  is  known  that 
37  pounds  of  tin  lose  5  pounds,  and  23  pounds  of 
lead  lose  2  pounds,  when  weighed  in  water.  How 
many  pounds  of  tin  and  of  lead  in  the  mass  ? 

58.  If  19  pounds  of  gold  lose  1  pound,  and  10  pounds  of 

silver  lose  1  pound,  when  weighed  in  water,  find  the 
amount  of  each  in  a  mass  of  gold  and  silver  weighing 
106  pounds  in  air  and  99  pounds  in  water. 

59.  Fifteen  sovereigns  should  weigh  77  pennyweights ;  but 

a  parcel  of  light  sovereigns,  having  been  weighed  and 
counted,  was  found  to  contain  9  more  than  was  sup- 
posed from  the  weight ;  and  it  appeared  that  21  of 
,  these  coins  weighed  the  same  as  20  true  sovereigns. 
How  many  were  there  altogether  ? 


60.  There  are  two  silver  cups,  and  one  cover  for  both.    The 

first  weighs  12  ounces,  and  with  the  cover  weighs 
twice  as  much  as  the  other  without  it ;  but  the  sec- 
ond with  the  cover  weighs  one-third  more  than  the 
first  without  it.     Find  the  weight  of  the  cover. 

61.  A  man  wishes  to  enclose  a  circular  piece  of  ground  with 

palisades,  and  finds  that  if  he  sets  them  a  foot  apart 
he  will  have  too  few  by  150 ;  but  if  he  sets  them  a 
yard  apart  he  will  have  too  many  by  70.  What  is  the 
circuit  of  the  piece  of  ground  ? 

62.  A  horse  was  sold  at  a  loss  for  $200 ;  but  if  it  had  been 

sold  for  $250,  the  gain  would  have  been  three-fourths 
of  the  loss  when  sold  for  $200.  Find  the  value  of 
the  horse. 

63.  A  and  B  shoot  by  turns  at  a  target.     A  puts  7  bullets 

out  of  '12,  and  B  9  out  of  12,  into  the  centre.  Be-- 
tween  them  they  put  in  32  bullets.  How  many  shots 
did  each  fire  ? 


PEOBLEMS.  149 


64.  A  boy  buys  a  number  of  apples  at  the  rate  of  5  for  2 

pence.  He  sells  half  of  them  at  2  a  penny  and  the 
rest  at  3  a  penny,  and  clears  a  penny  by  the  tran- 
saction.    How  many  does  he  buy  ? 

65.  A  person  bought  a  piece  of  land  for  ?  6750,  of  which 

he  kept  f  for  himself.  At  the  cost  of  $250  he  made 
a  road  which  took  ^^  of  the  remainder,  and  then  sold 
the  rest  at  12|-  cents  a  square  yard  more  than  double 
the  price  it  cost  him,  thus  clearing  his  outlay  and 
$500  besides.  How  much  land  did  he  buy,  and 
what  was  the  cost-price  per  yard  ? 

66.  A  boy  who  runs  at  the  rate  of  12  yards  per  second 

starts  20  yards  behind  another  whose  rate  is  10|- 
yards  per  second.  How  soon  will  the  first  boy  be 
10  yards  ahead  of  the  second  ? 

67.  A  merchant  adds  yearly  to  his  capital  one-third  of  it, 

but  takes  from  it,  at  the  end  of  each  year,  $  5000  for 
expenses.  At  the  end  of  the  third  year,  after  de- 
ducting the  last  $5000,  he  has  twice  his  original 
capital.     How  much  had  he  at  first  ? 

68.  A  shepherd  lost  a  number  of  sheep  equal  to  one-fourth 

of  his  flock  and  one-fourth  of  a  sheep  ;  then,  he  lost  a 
number  equal  to  one-third  of  what  he  had  left  and 
one-third  of  a  sheep  ;  finally,  he  lost  a  number  equal 
to  one-half  of  what  now  remained  and  one-half  a 
sheep,  after  which  he  had  but  25  sheep  left.  How 
many  had  he  at  first? 

69.  A  trader  maintained  himself  for  three  years  at  an  ex- 

pense of  $250  a  year ;  and  each  year  increased  that 
part  of  his  stock  which  was  not  so  expended  by  one- 
third  of  it.  At  the  end  of  the  third  year  his  original 
stock  was  doubled.     What  was  his  original  stock  ? 


150  ALGEBRA. 

70.  A  cask  contains  12  gallons  of  wine  and  18  gallons  of 

water ;  another  cask  contains  9  gallons  of  wine  and 
3  gallons  of  water.  How  many  gallons  must  be 
drawn  from  each  cask  to  produce  a  mixture  contain- 
ing 7  gallons  of  wine  and  7  gallons  of  water  ? 

71.  The  members  of  a  club  subscribe  each  as  many  dollars 

as  there  are  members.  If  there  had  been  12  more 
members,  the  subscription  from  each  would  have 
been  $10  less,  to  amount  to  the  same  sum.  Hdw 
many  members  were  there  ? 

72.  A  number  of  troops  being  formed  into  a  solid  square, 

it  was  found  there  were  60  men  over;  but  when 
formed  in  a  column  with  5  men  more  in  front  than 
before,  and  3  men  less  in  depth,  there  was  lacking 
one  man  to  complete  it.     Find  the  number  of  troops. 

73.  An  officer  can  form  the  men  of  liis  regiment  into  a  hol- 

low square  twelve  deep.  The  number  of  men  in  the 
regiment  is  1296.  Find  the  number  of  men  in  the 
front  of  the  hollow  square. 

74.  A  person  starts  from  P  and  walks  towards  Q  at  the 

rate  of  3  miles  an  hour ;  20  minutes  later  another 
person  starts  from  Q  and  walks  towards  P  at  the 
rate  of  4  miles  an  hour.  The  distance  from  P  to  Q 
is  20  miles.     How  far  from  P  will  they  meet  ?  > 

75.  A  person  engaged  to  work  a  days  on'' these  conditions : 

for  each  day  he  worked  he  was  to  receive  h  cents, 
and  for  each  day  he  was  idle  he  was  to  forfeit  c  cents. 
At  the  end  of  a  days  he  received  d  cents.  How 
many  days  was  he  idle  ? 

76.  A  banker  has  two  kinds  of  coins :  it  takes  a  pieces  of 

the  first  to  make  a  dollar,  and  h  pieces  of  the  second 
to  make  a  dollar.  A  person  wishes  to  obtain  c  pieces 
for  a  dollar.  How  many  pieces  of  each  kind  must 
the  banker  give  him  ? 


CHAPTER  XL 
Simultaneous  Equations  of  the  First  Degree. 

184.  If  one  equation  contain  two  unknown  quantities,  an 
indefinite  number  of  pairs  of  values  may  be  found  that  will 
satisfy  the  equation. 

Thus,  in  the  equation  a;  +  y  =  10,  any  values  may  be 
given  to  x,  and  corresponding  values  for  y  may  be  found. 
Any  pair  of  these  values  substituted  for  x  and  y  will  satisfy 
the  equation. 

185.  But  if  a  second  equation  be  given,  expressing  differ- 
ent relations  between  the  unknown  quantities,  only  one  pair 
of  Talues  of  x  and  y  can  be  found  that  will  satisfy  both 
equations. 

Thus,  if  besides  the  equation  a;  +  y  =  10,  another  equa- 
tion, x  —  y  =  2,  be  given,  it  is  evident  that  the  values  of 
a;  and  y  which  will  satisfy  both  equations  are 

for  6  +  4  =  10,  and  6  —  4  =  2;  and  these  are  the  only  val- 
ues of  X  and  y  that  will  satisfy  both  equations. 


186.  Equations  that  express  diffei^ent  relations  between 
the  unknown  quantities  are  called  independent  equations. 

Thus,  a; 4- y  =  10  and  x  —  y=2  are  independent  equa- 
tions ;  they  express  different  relations  between  x  and  y. 
But  x-\-y^  10  and  3a;4-3y=30  are  not  independent 


152  ALGEBRA. 

equations;  one  is  derived  immediately  from  the  other, 
and  both  express  the  same  relation  between  the  unknown 
quantities. 

187.  Equations  that  are  to  be  satisfied  by  the  same  val- 
ues of  the  unknown  quantities  are  called  simultaneous 
equations. 

188.  Simultaneous  equations  are  solved  by  combining 
the  equations  so  as  to  obtain  a  single  equation  containing 
only  one  unknown  quantity ;  and  this  process  is  called 
elimination. 

Three  methods  of  elimination  are  generally  given : 
I.    By  Addition  or  Subtraction. 
II.    By  Substitution. 
III.    By  Comparison. 

Elimination  by  Addition  or  Subtraction. 
(1)    Solve: 


2x 

-3y  = 

=  32/ 

(1) 

^x 

+  2y^ 

(2) 

Multiply  (1)  by  t 

2  and  (2)  by  3, 

4a;- 

6y  = 

8 

(3) 

9a;  + 

6.v  = 

96 

(4) 

Add  (3)  and  (4), 

13  a; 

104 

Substitute  the  value  of  x  in  (2), 

24 +  22/ =  32, 

•••  y  =  4. 

In  this  solution  y  is  eliminated  by  addition. 

(2)    Solve:  6;r  +  35y  =  177|  (1) 

8:i;-21y=    33  J  (2) 

Multiply  (1)  by  4  and  (2)  by  3, 

24x  +  140y  =  708  (3) 

24a;-    63 .v=    99  (4) 

Subtract  (4)  from  (3),  203  y  -  609 

.-.  y-3. 


SIMULTANEOUS   EQUATIONS.  153 

Substitute  the  value  of  y  in  (2). 
8a;-63  =  33. 
/.  a;  =12. 
In  this  solution  x  is  eliminated  by  subtraction. 

189.  Hence,  to  eliminate  an  unknown  quantity  by  addi- 
tion or  subtraction, 

Multiple/  the  equations  hy  such  numbers  as  will  make  the 
coefficients  of  this  unknown  quantity  equal  in  the  resulting 
equations. 

Add  the  resulting  equations,  or  subtract  one  from  the  other, 
according  as  these  equal  quantities  have  unlike  or  like  signs. 

Note.  It  is  generally  best  to  select  that  unknown  quantity  to  be 
eliminated  which  requires  the  smallest  multipliers  to  make  its  coeffi- 
cients equal ;  and  the  smallest  multiplier  for  each  equation  is  found 
by  dividing  the  L.  C.  M.  of  the  coefficients  of  this  unknown  quantity 
by  the  given  coefficient  in  that  equation.  Thus,  in  example  (2),  the 
L.  C.  M.  of  6  and  8  (the  coefficients  of  x),  is  24,  and  hence  the  smallest 
multipliers  of  the  two  equations  are  4  and  3  respectively. 

Sometimes  tbe  solution  is  simplified  by  first  adding  the 
given  equations,  or  by  subtracting  one  from  the  otber. 

(3)  x  +  ^9y=   51  (1) 

49a;  +  .    y=   99  (2) 

Add  (1)  and  (2),  50a;  +  503/=  150  (3) 

Divide  (3)  by  50,  x  +  f  =  3.  (4) 

Subtract  (4)  from  (1),  48  y  =  48, 

.-.  y  =  i. 

Subtract  (4)  from  (2),  48  a;  =  96, 

.-.  a;  =  2. 

Exercise  LXVIII. 
Solve  by  addition  or  subtraction : 

2x-}-Sy  =  7\    3.    7x  +  2y  =  S0)     5.    5:r  +  4y=58"l 

J  3a;  +  7?/=67J 


4^-5y  =  3J  y-Sx=   2)  3x  +  7y- 

x  —  2y  =  4:)     4.    3a;-5y  =  51|    6.   Sx  +  2y=S9) 
2x~    y  =  5j  2x  +  7y=^   3J  3y-2a;=13/ 


154  ALGEBRA. 


7.  3:r-42/  =  -5|  11.   12^;+    7y=176| 
4a;-5y=      IJ  S2/-19x=      3J 

8.  lla;  +  3y  =  100|  12.   2a;-7y=    81 

4a:-7y=     4J  4y-9^  =  19J 

9.  a;  +  49y  =  693)  13.    692/-17:c=    103  1 
J  14a;-13v  =  _41  J 


49a:+      2/  =  357J  14a;-13y 

17a;-f-3y=:571  14.    17^  +  302/  =  59 

16y  -  3a;  =  23  J  19a;  +  28y  =-  77 


Elimination  by  Substitution. 

(1)    Solve:  2a;  +  3y  =  8| 

3a;+7y=7J 

2a;  +  3y  =  8  (1) 

3a;  +  7y  =  V  (2) 

Transpose  3  y  in  (1),  2  a;  =  8  —  3  y. 

Divide  by  coefficient  of  re,      x=    ~    ^.  (4) 

Substitute  the  value  of  x  in  (2),  3  (    "^  '^)  +  '^y  =  7, 

24-9y  +  14y  =  14, 
5y  =  -10, 

.••y  =  -2. 

Substitute  the  value  of  y  in  (1),  2  a;  —  6  =  8, 

.-.  a;  =7. 

190.  Hence,  to  eliminate  an  unknown  quantity  by  sub- 
stitution, 

From  one  of  the  equations  obtain  the  value  of  one  of  the 
unknown  quantities  in  terms  of  the  other. 

Substitute  for  this  unknown  quantity  its  value  in  the  other 
equation,  and  reduce  the  resulting  equation. 


SIMULTANEOUS   EQUATIONS.  155 


Exercise  LXIX. 

Solve  by  substitution : 

1.   3:r-4y  =  2| 
7a;-9y=7J 

8. 

3a;-4y=18| 
3a;-|-2y=    0.' 

2.   7a;-5y-241 
4a:-3y  =  llJ 

9. 

9a;-5y  =  521 
8y-3a;=    8) 

3.     3a;  +  2y  =  32| 
20a; -3y-    li 

10. 

bx-2>y=   4| 
12y-7a;  =  10J 

4.   lla;-7y  =  37\ 
8a;  +  9y  =  4l/ 

11. 

9y-7a;  =  131 
15a;-7y=    9/ 

5.     7x+    5y  =  60| 
13a;-lly  =  10/ 

12. 

5a;-2y=    51  | 
19a;-3y  =  180/ 

6.   6a;-7y  =  42| 
7a;-6y=75J 

13. 

4a;+    9y  =  106| 
8a;  +  17y=198J 

' 

7.   10a;  +    9y  =  290| 
12a;-lly=130i 

14. 

8a;  +  3y  =  3| 
12a:  +  9y  =  3i 

- 

Elimination 

BY  Comparison. 

Solve:                         2  a;- 

-9y  = 

;'} 

3a;- 

-42/  = 

Transpose  dy  in  (1)  and  4?/  in 

>x-9y  =  ll, 
\x-^y  =  n. 
l(2),  2a;  =  11 +92/, 
3a;=7  +  4y. 

(1) 
(2) 
(3) 
(4) 

Divide  (3)  by  2  and  (4)  by  3, 

X      ^1+^^^ 
2 

(5) 

3 

(6) 

Equate  the  values  of; 

1 
^1 

[1+9.V      7+4.y 
2               3 

(V) 

166  ALGEBRA. 

Reduce  (7)  33  +  27y  =  U  +  S  y, 

192/ =  -19, 

.-.  2/  =  -l. 
Substitute  the  value  of  y  in  (1),  2  a;  +  9  =  11, 

.-.  x  =  l. 

191.  Hence,  to  eliminate  an  unknown  quantity  by  com- 
parison, 

From  each  equation  obtain  the  value  of  one  of  the  unknown 
quantities  in  terms  of  the  other. 

Form  an  equation  from  these  equal  values  and  reduce  the 
equation. 

Note.   If,  in  the  last  example,  (3)  be  divided  by  (4),  the  resulting 

equation,  -  = ^,  would,  when  reduced,  give  the  value  of  y. 

^  3       7  +  43/  ^  ^ 

This  is  the  shortest  method,  and  therefore  to  be  preferred. 

Exercise  LXX. 
Solve  by  comparison : 

1.  a:  +  15y  =  53|  8.   3y-7a:=   4| 

2.  4x+    9y  =  51|  9.   21y +  20a;  =  165  | 
8:;T-13y=    9)  77y- 30a:  =  295  i 

3.  4a;  +  3y  =  48|  10.    lla:-10y=141 
6y-3a;-22J  bx+    72/  =  41J 

4.  2a;  +  3y  =  43  1  11.    72/-3a;  =  139| 
10a;-    y=    7/  2x  +  b7/=    91 J 

5.  bx~    7y=    331  12.    17a:+122/=    59 1 
lla:+12y  =  100i  19a:-    4y  =  153J 

6.  5a-4-7y  =  43\  13.   24a:+    7y=    27 


} 


lla;+9y  =  69  J  8a;-33y=115 

8x-21ij=    33|  14.   a:  =  3y-19| 

6a:+35y  =  177i  y  =  3a;-23i 


SIMULTANEOUS   EQUATIONS.  157 

192.    Each  equation  must  be  simplified,  if  necessary,  be- 
fore the  elimination  is  performed. 

(1)    Solve:    (x~l)(y  +  2)  =  (x-S)(i/-l)  +  8)       . 
2:^-1      3(y-2)_.  [ 

5  4  J 

(x-l)(2/  +  2)  =  (x-3)(y-l)  +  8  (1) 

5  4  ^  ^ 

Simplify  (1),  xy  +  2x  —  y  -  2  =  xy  —  x  —  Sy  +  3  +  8. 

Transpose  and  combine,  3x  +  2y  =  13.  (3) 

Simplify  (2),  8  a;  -  4  -  loy  +  30  =  20. 

Transpose  and  combine,  8x  —  I5y  =  —  6.  (4) 

Multiply  (3)  by  8,  24:x  +  16y  =  104.  (6) 

Multiply  (4)  by  3,  24a;  -  45t/  =  -  18.  (6) 

Subtract  (6)  from  (5),  fily  =  122, 

.'.y  =  2. 
Substitute  the  value  of  y  in  (3),  3  a;  +  4  =  13, 

.-.a;  =  3. 


Exercise  LXXI. 
Solve : 

1.  .:(y  +  7)  =  2/(a;+l))  3.    _2     =     ^ 
2x-i-20  =  Sy  +  l      J  ^+^      2/-2 

5(:r+3)-3(y-2)4-2 

2.  2:^-^:^-4  =  0]        4.  £zii_^i:2_Q. 

5  [  5  10  ( 

3y+^-9  =  oJ  ^+.!iz:2_3      J 

^  6         4 

5.    (:r+l)(y  +  2)-(ar  +  2)(y  +  l)  =  -n 
3(:r+3)-4(3/  +  4)==-8  J 

6    ^-2  10-:y_y-10 

5  3               4 

2y  +  4  2x  +  y_x  +  lS 

3  8               4 


158 


ALGEBRA. 


7    ^-+1      y-\-2_2{x-y)  ' 

3  4  5 

— : ^     —  ^y  —  X 

4  3-^ 


g^    Sx-2y^5x~Sy^^^^  . 
5  3 

2x  —  Sy  ,  Ax  —  3y  ,  ^ 


15. 


16. 


5  10 

6         4 


3:r  +  12y_g 

11 
l-3a:_ll-3.y 

7  5 


2x-y+3      x-2y+S_^ 

3  4 

3a;-4y+3      4a;-27/-9_, 

4  "^  3 


10.   l|a;=143/  +  43^| 


U. 


13 


•«s 


V'5 


3  19 


17.    5:r-i(52/  +  2)  =  32") 
3y  +  i(a;  +  2)=9      i 

5.    3a;-.25y=28     | 
.12a:+.7y=2.54j 


2      6:r-5y  +  4      3a7  +  22/+l 


12.   ^±y  =  15 

y  —  X      8 

9^_^_iM  =  100 

7 

Q      ^  — 2y_^.r  ,  y 
4  2  '  3 


19.    7(:r-l)  =  3(3/  +  8) 

4:g  +  2^5.v+9 

9  2 


20.    7:r  +  i(2y+4)  =  16| 


3y-i(:r  +  2)=8 


14. 


4a;-3y-7_,3rg_2?/_5 


5  10      15      6 


2     20 


15 


6      10  J 


O 


SIMULTANEOUS    EQUATIONS. 


159 


21. 


22. 


23. 


24. 


13       ^ 


42/-2 


5a7+6y      3a;  — 2y^o 
6  4  -^ 


bx 


3      3^-19 


2  2 

2x-{-y     9a:— 7 
2 


4- 


3r 


3(.V  +  3) 


8 


4a:-{-  5y 
16 


3y+ii=^^-y(^+^y)+3i-4^ 

x  —  y-\-^ 
(a:+7)(y-2)  +  3  =  2ary-(y-l)(a;H-l) 

6a;  +  9  ,  3a7  +  5y_Qi  i  3a7  +  4  ^ 
-T~+4^-6  -^*  +  ~^~ 
8y+7  ■  6a;-3y_.      4y-9 
10  2y-8  "^      5 


25.    07- 

2/  + 


2y~x 
23 -a; 

y-3  . 

a-- 18 


20 


59 -2a; 


3Q__73-3y 
3 


Literal  Simultaneous  Equations. 

193.    The  method  of  solving  literal  simultaneous  equa- 
tions is  as  follows : 

Solve :  ax-\-hy=^7n\ 

=  71  / 


cx-\-  dy=^ 

ax  +  by  =  m 
ex  +  dy  =  n 
Multiply  (1)  by  c,  acx  +  hey  =  em 

Multiply  (2)  by  a,  acx  +  ady  =  an 

Subtract  (4)  from  (3), 


(1) 
(2) 
(3) 
(4) 


(6c  —  ad)  y  =  cm  —  an 


Divide  by  coefficient  of  y, 


y 


be  —  ad 


160 


ALGEBRA. 


To  find  the  value  of  x 

Multiply  (1)  by  d. 
Multiply  (2)  by  6, 
Subtract  (6)  from  (5), 


adx  +  hdy  =  dm  (5) 

hex  +  hdy  =  hn  (6) 

{ad  —bc)x=  dm  —  hn 

dm  —  hn 


Divide  by  coefficient  oi  x,  ~  ,     , 

"^  ad —he 

Exercise  LXXII. 
Solve : 

1.    x-[-y  =  a\        3.   mx -\-ny  =  a'\     5.   mx  —  ny=r    \ 
b  J  px  -}-  gy  =  b    J  m'x  +  n'y  =  /  J 


x~y 


2.    ax-\~by  =  c\     4.   ax-{-by  =  e\       6.    ax-j-by^c} 
px-{-  qy  =  r  )  ax-{-cy^^d)  dx  -\-fy  =  c^  i 


7.   ?  +  |  =  e 
a      0 

X  ^  y .____ 
b   '  a 


12. 


^-.y  +  1 

x-y-l 
x  +  y  +  l 
x-\-y-l 


8.    abx  -f-  cdy  =  2 
d-b 


13.    ax  =  by  +  ^^ 
(a—b)x  =  (a  +  b)y 


9. 


b-\-y      Sa-\-x 
ax  -{-  2by  =  d 


14.    ax  +  5?/ 


5  +  2/      a  +  ^ 


10. 


.__X 


1 


a  +  5      a  —  5      a-\-b 

X      ■      y     —     1 
a  +  6      a  —  5      a~b  ^ 


15. 


f--^  =  2al 


X  —  y  __  rr  +  y 
2a6  ~aM^'2 


11.    a(a  — a:)  =  5(a:+y  — a)  I    16.    Jo;  —  Z)C  =  ay  —  ac  1 
a(y  —  b—x)=b(y  —  b)}  x  —  y  =  a  —  b         J 


SIMULTANEOUS    EQUATIONS.  161 


17.    ~  =  c 

y-b 

a  (x  —  a) -{- b  (y  —  b) -{-  abc  =  6 

1&    (a-i-b)x  —  (a  —  b)y  =  4iab  ) 

(a-b)x  +  (a  +  b)y  =  2a'-2b')    . 

19.  (x+a)(y+b)-(x-a'){2j-b)  =  2(a~byi 
x-y+2{a-h)=0  J 

20.  {a  +  b){x  +  y)-{a~b){x-y)  =  a'\ 
{a-b){x  +  y)  +  {a  +  b){x~y)  =  b'\ 

194.  Fractional  simultaneous  equations,  of  which  the  de- 
nominators are  simple  expressions  and  contain  the  unknown 
quantities,  may  be  solved  as  follows : 

(1)    Solve:  -  +  -  =  7^1  ^ 

X     y 

c   ,  d 
-  +  -=:n 
X     y 

a     b  ,^^ 

-  +  -  =  m.  (1) 

'-  +  ^  =  n.  (2) 

Multiply  (1)  by  c,  ^'  +  ^  =  cm.  (3) 

X       y 

Multiply  (2)  by  a,  ^  +  ^  =  an.  (4) 

X       y 

Subtract  (4)  from  (3),  ^c  -  ad  _^^_^^ 

y 

Multiply  both  sides  by  y,     he  —  ad=  {cm  —  an)  y, 

he  —  ad 

.-.?/  = . 

cm  — an 

Multiply  (1)  by  cZ,  ^l+^JL^^dm.  (5) 

X        y  ^  ' 

Multiply  (2)  by  6.  t  +  ^==bn.  (6) 

X       y  ^  ' 


162 


ALGEBRA. 


Subtract  (6)  from  (5), 


ad—  he 


dm  —  hn. 


Multiply  both  sides  by  x,     ad—hc  =  {dm  —  bn) x, 

ad— be 


dm  —  bn 


(2)    Solv.e: 


Sx      by 


_7_ 
%x 


Multiply  (2)  by  4, 
Add  (1)  and  (3), 
Divide  both  sides  by  19, 


1 
lOy 


Zx      by 

6a;      lOy 

^± — ^=12. 
2>x      by 

2>x 


2>x 
.'.x  =  \ 


Substitute  the  value  of  x  in  (1), 

Transpose, 

Divide  both  sides  by  2, 


5+^=7. 
by 

by 


Solve 


.1+?= 

.10  1 

9          ^ 
3.    -_   5 

X      y 

X      Zy 

'-  +  '-  = 

.20 

r+- 

X      y 

4:x      y 

Exercise   LXXIII. 

_4 

27 

n 

72 


y 

5  =  6 

y 


2.      i  +  ?:=:a 

X      y 
X      y 


1      2 
X      y 


=  4 


6. 


a      5 «/? 

X      y       b 


(1) 
(2) 
(3) 


SIMULTANEOUS   EQUATIONS. 


163 


7.1  +  ^^5 
ax      by 


5^ 

ax 


2_ 
by 


8.   ^  +  ^ 
nx      my 

n      m 

X      y~ 


f.    a  ,  h 

X      y 

b      a 
=  n 

X     y 


195.  If  three  simultaneous  equations  are  given,  involv- 
ing three  unknown  quantities,  one  of  the  unknown  quanti- 
ties must  be  eliminated  between  two  pairs  of  the  equations  ; 
then  a  second  between  the  resulting  equations. 

196.  Likewise,  if  four  or  more  equations  are  given, 
involving  four  or  more  unknown  quantities,  one  of  the 
unknown  quantities  must  be  eliminated  between  three  or 
more  pairs  of  the  equations  ;  then  a  second  between  the 
pairs  that  can  be  found  of  the  resulting  equations ;  and 
so  on. 


)lve:                     2x 
Sx 
bx 

-3y-f42=    4-) 
+  5y-7z  =  12  [ 
-    y~8z=^   5) 

(1) 
(2) 
(3) 

imirate  z  between  two 

pairs  of  these  equations. 

Multiply  (1)  by  2, 

(3)i3 

4a;-6y  +  82=    8 
5a;-    y-82=    5 

(4) 

Add, 

9x~7y          =13 

(5) 

Multiply  (1)  by  7, 
Multiply  (2)  by  4, 
Add, 

14a;-21y  +  282  =  28 
12a; +  20?/ -282  =  48 
26a;-      y             =76 

(6) 

Multiply  (6)  by  7, 

(5)  is 

Subtract  (5)  from  (7), 

182  a; -7?/ =  532 
9a;-7.v=    13 
173  a;           =519 
•     r  —  3 

(7) 

Substitute  the  value 

.  .    X  —  o. 

of  X  in  (6),                    78  -  2/  = 

=  76, 
=  2. 
=  4, 
=  1. 

Substitute  the  values  of  x  and  y  in  (1),  6  —  6  +  42  = 

.•.  2  = 

164  ALGEBRA. 


^  ,  Exercise  LXXIV. 

Solve : 

'l.   6x  +  Sy-6z  =  4:  ■\  10.   Sx-y  +  z  =  17 

x-2y  +  2z  =  2     J  7a;  +  4y-5s  =  3 

2.   4a;  — 5y  +  2z==6     -j  11.   x^9/  +  z  =  b 

2a;  +  3y  — 2  =  20     \  Sa^  —  Sy  +  7z  =  75 


7^_4y_|.32  =  35  J  9^-112  +  10  =  0 

3.   x  +  i/  +  z  =  Q  )        12.   ^  +  2?/  +  3;^  =  6 


5:r  +  4y  +  3z  =  22       V  2:^  +  4y  +  2z  =  8 

15a;  +  107/  +  6z  =  53  J  Sx +  27/  +  8z  =  10l  J 

4.  45;-3y  +  2;  =  9     ^  13.   a:-3y-22;  =  l 

9a;  +  y-5z  =  16   I  2rc- 3y  +  5^  =  - 19 

a;-4y  +  3z  =  2     ]  5a;  +  2y-z  =  12 

5.  8a;  +  4y-32  =  6)  14.    3:r-2y==5 


^  +  3y_;2  =  7        ^  4;^_3y  +  2z  =  ll 

4a;-5y  +  4z  =  8  ]  .'r-2y-5z  =  -7 

6.    12a;  +  5y-4z  =  29  ^  15. 

=  58   I 


13a;-2y  +  5z  =  58 

17rr  — y  — z  =  15  J 

7.  y-a;  +  z  =  -5  -j  16.  2a;-3y  =  3 
2_y_^  =  _25  I  3y-4z=7 
a;  +  y  +  z  =  35       j  4z-5.7;  =  2 

8.  :r  +  y  +  z  =  30  ^  17.    3a;-4y  +  6z  =  l 
8a:  +  4y-f  22  =  50  \  2x  +  2y-z  =  l 


27a;  +  9y  +  32  =  64  j  7:^-6y  +  72  =  2 

9.    15y  =  242-10a;  +  41    ^  18.   7:r-3y  =  30 
15a;  =  12y-16z  +  10  9y-52  =  34 

18a:-(7z-13)  =  14y  J  a:  +  y  +  2  =  33 


SIMULTANEOUS    EQUATIONS. 


165 


19. 

^+M=^ 

y+|+|=- 

^+1+1-^^ 

20. 

l+?=5  ] 
X     y 

?_4_6 

y    z 

3-4  =  5 

0      a;              J 

21. 

1+1-1=. 

1-1+1=5 

X      y      z 

11  1_^ 
y     z      X 

22. 

hz-{-cy=^a   ' 

az-{-cx  =  h     ■' 

ay  +  bx  =  c  . 

23.    ?_A  I  i  =  7| 

a;      5y      z 

^  +  ;^+^-10i 
da;      zy      z 

A_JL  '  ^  =  16-1 
5:r      22/"^z  '^ 


24.   ?_?  +  ^  =  2.9 


10.4 


5_6_7 
X      y      z 

9  +  L0_8^14.9 
y       z       a; 


25.    2  +  1. 

-3  =  0 

^     y 

z 

3     2 

-2  =  0 

^     y 

1  ,  1 

— I 

-t=o 

X      z 

3 

26.    aa:  +  5y  +  <^z  =  a 
aa:  —  hy  —  cz=h 

ax  -j-  cy  -\-  bz  =  c 


97    ^^~y  —  ^y  4-  ^^ a;  —  y 

3      ~     ^  5 


4 


28.  ^-y -y~^ 

a  b 


X  —  a  —  b 

a-\-b-\-  c 


*  Subtract  from  the  sum  of  the  three  equations  each  equation  separately, 
f  Multiply  the  equations  by  a,  h,  and  c,  respectively,  and  from  the  sum 
of  the  results  subtract  the  double  of  each  equation  separately. 


CHAPTER  XIL 

Problems  producing  Simultaneous  Equations. 

197.  It  is  often  necessary  in  the  solution  of  problems 
to  employ  two  or  more  lette^^s  to  represent  the  quantities  to 
be  found.  In  all  cases  the  conditions  must  be  sufficient 
to  give  just  as  many  equations  as  there  are  unknown  quan- 
tities employed. 

If  there  be  Tuore  equations  than  unknown  quantities, 
some  of  them  are  superfluous  or  contradictory  ;  if  there  be 
less  equations  than  unknown  quantities,  the  problem  is  in- 
determinate or  impossible. 

(1)  When  the  greater  of  two  numbers  is  divided  by  the 
less  the  quotient  is  4  and  the  remainder  3 ;  and  when 
the  sum  of  the  two  numbers  is  increased  by  38,  and 
the  result  divided  by  the  greater  of  the  two  numbers, 
the  quotient  is  2  and  the  remainder  2.  Find  the 
numbers. 

Let  X  =  the  greater  number, 

and  y  =  the  smaller  number. 

x-Z      . 


Then 


y 


and        ^+.V  +  38-2^,^ 

X 

From  the  solution  of  these  equations  x  ==  47,  and  3/  =  11. 

(2)  If  A  give  B  $10,  B  will  have  three  times  as  much  money 
as  A.  If  B  give  A  $10,  A  will  have  twice  as  much 
money  as  B.     How  much  has  each  ? 


PROBLEMS.  167 


Let  X  =  number  of  dollars  A  has, 

and  y  =  number  of  dollars  B  has. 

Then  y  +  10  =  number  of  dollars  B  has,  and  a;  —  10  --  number 
of  dollars  A  has  after  A  gives  .|  10  to  B.. 

.-.  y  +  10  =  3  (x  -  10),  and  a;  +  10  =  2  (y  -  10). 

From  the  solution  of  these  equations,  a;  -^  22  and  y  =  26. 

Therefore,  A  has  $22  and  B  $26. 

Exercise  LXXV. 

i.  The  sum  of  two  numbers  divided  by  2  gives  as  a  quo- 
tient 24,  and  the  difference  between  them  divided  by 
2  gives  as  a  quotient  17.     What  are  the  numbers  ? 

2.  The  number  144  is  divided  into  three  numbers.  When 

the  first  is  divided  by  the  second,  the  quotient  is  3 
and  the  remainder  2  ;  and  when  the  third  is  divided 
hj  the  sum  of  the  other  two  numbers,  the  quotient  is 
2  and  the  remainder  6.     Find  the  numbers. 

3.  Three  times  the  greater  of  two  numbers  exceeds  twice 

the  less  by  10 ;  and  twice  the  greater  together  with 
three  times  the  less  is  24.     Find  the  numbers. 

4.  If  the  smaller  of  two  numbers  be  divided  by  the  greater, 

the  quotient  is  .21  and  the  remainder  .0057;  but  if 
the  greater  be  divided  by  the  smaller,  the  quotient 
is  4  and  the  remainder  .742.  What  are  the  num- 
bers ?  .  ■•'.'■■ 

5.  Seven  years  ago  the  age  of  a  father  was  four  times  that 

of  his  son ;  seven  years  hence  the  age  of  the  father 
will  be  double  that  of  the  son.    What  are  their  ages? 

6.  The  sum  of  the  ages  of  a  father  and  son  is  half  what  it 

will  be  in  25  years ;  the  difference  between  their 
ages  is  one-third  of  what  the  sum  will  be  in  20  years. 
What  are  their  ages  ? 


168  ,  ■     _  ALGEBRA. 


7.  If  B  give  A  $25  they  will  have  equal  sums  of  money; 

but  if  A  give  B  $22,  B's  money  will  be  double  that 
of  A's.     How  much  has  each  ? 

8.  A  farmer  sold  to  one  person  30  bushels  of  wheat  and 

40  bushels  of  barley  for  $67.50;  to  another  persorv 
he  sold  50  bushels  of  wheat  and  30  bushels  of  barley 
for  $85.  Whai  was  the  price  of  the  wheat  and  of 
the  barley  per  bushel  ? 

9.  If  A  give  B  $5  he  will  then  have  $6  less  than  B  ;  but 

if  he  receive  $5  from  B,  three  times  his  money  will 
be  $20  more  than  four  times  B's.  How  much  has 
each  ? 

10.  The  cost  of  12  horses  and  14  cows  is  $1900;  the  cost 
of  5  horses  and  3  cows  is  $  650.  What  is  the  cost  of 
a  horse  and  a  cow  respectively  ? 

Note  I.   A  fraction  of  which  the  terms  are  unknown  may  be  rep- 
resented by  -• 

Ex.  A  certain  fraction  becomes  equal  to  ^  if  3  be  added  to 
its  numerator,  and  equal  to  |-  if  3  be  added  to  its  de- 
nominator.    Determine  the  fraction. 
Let  ^  =  the  required  fraction. 

By  the  conditions  ^        =  |, 

y 

and  "^ 


2/  +  3       ' 

From  the  solution  of  these  equations  it  is  found  that 
x=6, 
y  =  18. 
Therefore  the  fraction  =  -f^. 

11.  A  certain  fraction  becomes  equal  to  2  when  7  is  added 
to  its  numerator,  and  equal  to  1  when  1  is  subtracted 
from  its  denominator.     Determine  the  fraction. 


PROBLEMS.  169 


12.  A  certain  fraction  becomes  equal  to  -|-  when  7  is  added 
to  its  denominator,  and  equal  to  2  Avhen  13  is  added 
to  its  numerator.     Determine  the  fraction. 

13.  A  certain  fraction  becomes  equal  to  -J  when  the  denom- 
inator is  increased  by  4,  and  equal  to  J^  when  the 
numerator  is  diminished  by  15.  Determine  the  frac- 
tion. 

14.  A  certain  fraction  becomes  equal  to  -f  if  7  be  added  to 
the  numerator,  and  equal  to  f  if  7  be  subtracted  from 
the  denominator.     Determine  the  fraction. 

15.  Find  two  fractions  with  numerators  2  and  5  respectively, 
whose  sum  is  1^,  and  if  their  denominators  are  inter- 
changed their  sum  is  2. 

16.  A  fraction  which  is  equal  to  f  is  increased  to  -^j  when 
^^  )h.-t^  '■       a  certain  number  is  added  to  both  its  numerator  and 

denominator,  and  is  diminished  to  J-  when  one  more 
than  the  same  number  is  subtracted  from  each.     De- 
termine the  fraction. 
Note  II.    A  number  consisting  of  two  digits  which  are  unknown 
may  be  represented  by  10  a;  +  y,  in  which  x  and  y  represent  the  digits 
of  the  number.     Likewise,  a  number  consisting  of  three  digits  which 
are  unknown  may  be  represented  by  100  a;  +  lOy  +  z,  in  which  x,  y, 
and  z  represent  the  digits  of  the  number. 

For  example,  consider  any  number  expressed  by  three  digits,  as 
364.  The  expression  3G4  means  300  +  GO  +  4  ;  or,  100  times  3  +  10 
times  G  +  4. 

Ex.  The  sum  of  the  two  digits  of  a  number  is  8,  and  if  36 
be  added  to  the  number  the  digits  will  bo  inter- 
changed.    What  is  the  number  ? 

Let  X  =  the  digit  in  the  tens'  place, 

and  y  =  the  digit  in  the  units'  place. 

Then  lOx  +  y  =  the  number. 

By  the  conditions,     a;  +  3/  =  8,  (1) 

and  10x  +  y  +  3Q  =  l0yi-x.  (2) 


170  ALGEBEA. 


From  (2),  9a;-9t/  =  -36. 

Divide  by  9,  x  —  ?/  =  —  4. 

Add  (1)  and  (3),  2x  =  i, 

.-.a;  =  2. 
Subtract (3) from  (1),  2y  =  12, 

.•.y  =  6. 

Hence,  the  number  is  26. 

17.  The  sum  of  the  two  digits  of  a  number  is  10,  and  if  54 

be  added  to  the  number  the  digits  will  be  inter- 
changed.    What  is  the  number  ? 

18.  The  sum  of  the  two  digits  of  a  number  is  6,  and  if  the 

number  be  divided  by  the  sum  of  the  digits  the  quo- 
tient is  4.    What  is  the  number  ? 

19.  A  certain  number  is  expressed  by  two  digits,  of  which 

the  first  is  the  greater.  If  the  number  be  divided  by 
the  sum  of  its  digits  the  quotient  is  7  ;  if  the  digits 
be  interchanged,  and  the  resulting  number  diminished 
by  12  be  divided  by  the  difference  between  the  two 
digits,  the  quotient  is  9.     What  is  the  number  ? 

20.  If  a  certain  number  be  divided  by  the  sum  of  its  two 

digits  the  quotient  is  6  and  the  remainder  3 ;  if  the 
digits  be  interchanged,  and  the  resulting  number  be 
divided  by  the  sum  of  the  digits,  the  quotient  is  4 
and  the  remainder  0.    What  is  the  number  ? 

21.  If  a  certain  number  be  divided  by  the  sum  of  its  two 

digits  diminished  by  2,  the  quotient  is  5  and  the  re- 
mainder 1  ;  if  the  digits  be  interchanged,  and  the 
resulting  number  be  divided  by  the  sum  of  the  digits 
increased  by  2,  the  quotient  is  5  and  the  remainder 
8.    Find  the  number, 

22.  The  first  of  the  two  digits  of  a  number  is,  when  doubled, 

3  more  than  the  second,  and  the  number  itself  is  less 
by  6  than  five  times  the  sum  of  the  digits.  What 
is  the  number  ? 


PROBLEMS.  171 


23.  A  number  is  expressed  by  three  digits,  of  which  the  first 

and  last  are  alike.  By  interchanging  the  digits  in 
the  units'  and  tens'  places  the  number  is  increased  by 
54  ;  but  if  the  digits  in  the  tens'  and  hundreds'  places 
are  interchanged,  9  must  be  added  to  four  times  the 
resulting  number  to  make  it  equal  to  the  original 
number.     What  is  the  number  ? 

24.  A  number  is  expressed  by  three  digits.    The  sum  of  the 

digits  is  21 ;  the  sum  of  the  first  and  second  exceeds 
the  third  by  3 ;  and  if  198  be  added  to  the  number, 
the  digits  in  the  units'  and  hundreds'  places  will  be 
interchanged.     Find  the  number. 

25.  A  number  is  expressed  by  three  digits.     The  sum  of 

the  digits  is  9 ;  the  number  -is  equal  to  forty-two 
times  the  sum  of  the  first  and  second  digits  ;  and  the 
third  digit  is  twice  the  sum  of  the  other  two.  Find 
the  number. 

26.  A  certain  number,  expressed  by  three  digits,  is  equal  to 

forty-eight  times  the  sum  of  its  digits.  If  198  be 
subtracted  from  the  number,  the  digits  in  the  units' 
and  hundreds'  places  will  be  interchanged ;  and  the 
sum  of  the  extreme  digits  is  equal  to  twice  the  mid- 
dle digit.     Find  the  number. 

Note  III.  If  a  boat  move  at  the  rate  of  x  miles  an  hour  in  still 
water,  and  if  it  be  on  a  stream  that  runs  at  the  rate  of  y  miles  an 
hour,  then 

X  +y  represents  its  rate  down  the  stream, 
x  —  y  represents  its  rate  up  the  stream. 

27.  A  waterman  rows  30  miles  and  back  in  12  hours.     He 

finds  that  he  can  row  5  miles  with  the  stream  in  the 
same  time  as  3  against  it.  Find  the  time  he  was 
rowing  up  and  down  respectively. 


172  ALGEBRA. 


28.  A  crew,  whicli  can  pull  at  the  rate  of  12  miles  an  hour 

down  the  stream,  finds  that  it  takes  twice  as  long  to 
come  up  the  river  as  to  go  down.  At  what  rate  does 
the  stream  flow  ?       ' 

29.  A  man  sculls  down  a  stream,  w^hich  runs  at  the  rate  oi 

4  miles  an  hour,  for  a  certain  distance  in  1  hour  and 
40  minutes.  In  returning  it  takes  him  4  hours  and 
15  minutes  to  arrive  at  a  point  3  miles  short  oi  his 
starting-place.  Find  the  distance  he  pulled  down 
the  stream  and  the  rate  of  his  pulling. 

30.  A  person  rows  down  a  stream  a  distance  of  20  miles 

and  back  again  in  10  hours.  He  finds  he  can  row 
2  miles  against  the  stream  in  the  same  time  he  can 
row  3  miles  with  it.  Find  the  time  of  his  rowing 
down  and  of  his  rowing  up  the  stream ;  and  also  the 
rate  of  the  stream. 

Note  IV.  When  commodities  are  mixed,  it  is  to  be  observed  that 
the  quantity  of  the  mixture  =  the  quantity  of  the  ingredients ;  the 
cost  of  the  mixture  =  the  cost  of  the  ingredients. 

Ex.  A  wine-merchant  has  two  kinds  of  wine  which  coct 
72  cents  and  40  cents  a  quart  respectively.  How 
much  of  each  must  he  take  to  make  a  mixture  of  50 
quarts  worth  60  cents  a  quart? 

Let  X  =  required  number  of  quarts  worth  72  cents  a 

quart, 
and  y  =  required  number  of  quarts  worth  40  cents  a 

quart. 
Then,         72  a;  =  cost  in  cents  of  the  first  kind, 

AQy  ==  cost  in  cents  of  the  second  kind  of  wine, 
and  3000  =  cost  in  cents  of  the  mixture. 

.-.  x  +  y  =  bO, 
72  x  + 403/  =  3000. 
From  which  equations  the  values  of  x  and  y  may  be  found. 


PROBLEMS  173 


31.  A  grocer  mixed  tea  that   cost  him  42  cents  a  pound 

with  tea  that  cost  him  54  cents  a  pound.  He  had  30 
pounds  of  the  mixture,  and  by  selling  it  at  the  rate 
of  60  cents  a  pound,  he  gained  as  much  as  10  pounds 
of  the  cheaper  tea  cost  him.  How  many  pounds  of 
each  did  he  put  into  the  mixture  ? 

32.  A  grocer  mixes  tea  that   cost  him  90  cents  a  pound 

with  tea  that  cost  him  28  cents  a  pound.  The  cost 
of  the  mixture  is  $61.20.  He  sells  the  mixture  at 
50  cents  a  pound,  and  gams  $3.80.  How  many 
pounds  of  each  did  he  put  into  the  mixture  ? 

33.  A  farmer  has  28  bushels  of  barley  worth  84  cents  a 

bushel.  With  his  barley  he  wishes  to  mix  rye  worth 
$1.08  a  bushel,  and  wheat  worth  $1.44  a  bushel,  so 
that  the  mixture  may  be  100  bushels,  and  be  worth 
$1.20  a  bushel.  How  many  bushels  of  rye  and  of 
wheat  must  he  take  ? 

Note  V.  It  is  to  be  remembered  that  if  a  person  can  do  a  piece 
of  work  in  x  days,  the  part  of  the  work  he  can  do  in  one  day  will  be 
represented  by  I. 

Ex.  A  and  B  together  can  do  a  piece  of  work  in  48  days ; 
A  and  C  together  can  do  it  in  30  days ;  B  and  C  to- 
gether can  do  it  in  26-|  days.  How  long  will  it  take 
each  to  do  the  work  ? 

Let  X  =  the  number  of  days  it  will  ta^e  A  alone  to  do  the  work, 

y  =  the  number  of  days  it  will  take  B  alone  to  do  the  work, 

and      z  =  the  number  of  days  it  will  take  C  alone  to  do  the  work. 

Then,  _,  _,  _,  respectively,  will  denote  the  part  each  can  do 
X    V    z  ■         J 

^  in  a  day, 

and  -  +  -  will  denote  the  part  A  and  B  together  can  do  in  a  day, 
X     y 

but  —  will  denote  the  part  A  and  B  together  can  do  in  a  day- 


174  ALGEBRA. 

Therefore,  1  +  1  =  _L  (1) 

a;     y     48  ^  ^ 

Likewise,  1  + 1  =  Ji  (2) 

and  1  +  1  =  Jl  =  A  (3^ 

y     z      26|      80  ^^ 

Add  (1),  (2),  and  (3),        ?  +  ?  +  ?     ^Ji  (4) 
X     y     z          120 

Multiply  (1)  by  2,  ?  +  ?  =2.  (5) 

Subtract  (5)  from  (4), 


Subtract  the  double  of  (2)  from  (4),  -  ==  77^ 
Subtract  the  double  of  (3)  from  (4), 

34.  A  and  B  together  earn  ?40  in  6  days;  A  and  C  to- 

gether earn  $54  in  9  days;  B  and  C  together  earn 
§80  in  15  days.     What  does  each  earn  a  day? 

35.  A  cistern  has  three  pipes,  A,  B,  and  C.     A  and  B  will 

fill  it  in  1  hour  and  10  minutes ;  A  and  C  in  1  hour 
and  24  minutes ;  B  and  C  in  2  hours  and  20  minutes. 
How  long  will  it  take  each  to  fill  it  ? 

36.  A  warehouse  will  hold  24  boxes  and  20  bales ;  6  boxes 

and  14  bales  will  fill  half  of  it.  How  many  of  each 
alone  will  it  hold  ? 

37.  Two  workmen  together  complete  some  work  in  20  days ; 

but  if  the  first  had  worked  twice  as  fast,  and  the  sec- 
ond half  as  fast,  they  would  have  finished  it  in  15  days. 
How  long  would  it  take  each  alone  to  do  the  work  ? 

38.  A  jDurse  holds  19  crowns  and  6  guineas ;  4  crowns  and 

5  guineas  fill  -J-J  of  it.  How  many  of  each  alone  will 
it  hold  ? 


PROBLEMS.  175 


39.  A  piece  of  work  can  be  completed  by  A,  B,  and  0  to- 

gether in  10  days;  by  A  and  B  together  in  12  days; 
by  B  and  C,  if  B  work  15  days  and  C  30  days.  How 
long  will  it  take  each  alone  to  do  the  work? 

40.  A  cistern  has  three  pipes,  A,  B,  and  0.     A  and  B  will 

fill  it  in  a  minutes ;  A  and  C  in  5  minutes ;  B  and 
C  in  c  minutes.  How  long  will  it  take  each  alone  to 
fill  it? 

Note-  VI.  In  considering  the  rate  of  increase  or  decrease  in  quan- 
tities, it  is  usual  to  take  100  as  a  common  standard  of  reference,  so 
that  the  increase  or  decrease  is  calculated  for  every  100,  and  there- 
fore called  per  cent. 

It  is  to  be  observed  that  the  representative  of  the  number  result- 
ing after  an  increase  has  taken  place  is  100  +  increase  per  cent ;  and 
after  a  decrease,  100  —  decrease  per  cent. 

Interest  depends  upon  the  time  for  which  the  money  is  lent,  as 
well  as  upon  the  rate  per  cent  charged ;  the  rate  per  cent  charged 
being  the  rate  per  cent  on  the  principal  for  one  year.     Hence, 

c,-      1    •   .        .       Principal  X  Rate  X  Time 

Simple  interest  == ■ — , 

^  100 

where  Time  means  number  of  years  or  fraction  of  a  year. 

Amount  =  Principal  +  Interest. 

In  questions  relating  to  stocks,  100  is  taken  as  the  representative 
of  the  stock,  the  price  represents  its  market  value,  and  the  per  cent 
represents  the  interest  which  the  stock  bears.  Thus,  if  six  per  cent 
stocks  are  quoted  at  lOS,  the  meaning  is,  that  the  price  of  $100  of 
the  stock  is  $108,  and  that  the  interest  derived  from  $100  of  the 
stock  will  be  yf  q-  of  $100,  that  is,  $6  a  year.  The  rate  of  interest  on 
the  money  invested  will  be  \^^  of  G  per  cent. 

41.  A  man  has  $  10,000  invested.    For  a  part  of  this  sum  he 

receives  5  per  cent  interest,  and  for  the  rest  4  per 
cent ;  the  income  from  his  5  per  cent  investment  is 
$50  more  than  from  his  4  per  cent.  How  much  has 
he  in  each  investment  ? 


176  ALGEBRA. 


42.  A  sum  of  money,  at  simple  interest,  amounted  in  6 

years  to  §26,000,  and  in  10  years  to  $30,000.  Find 
the  sum  and  the  rate  of  interest. 

43.  A  sum  of  money,  at  simple  interest,  amounted  in  10 

months  to  1 26,250,  and  in  18  months  to  $27,250. 
Find  the  sum  and  the  rate  of  interest. 

44.  A  sum  of  money,  at  simple  interest,  amounted  in  ini 

years  to  a  dollars,  and  in  n  years  to  h  dollars.  Find 
the  sum  and  the  rate  of  interest. 

45.  A  sum  of  money,  at  simple  interest,  amounted  in  a 

months  to  c  dollars,  and  in  h  months  to  d  dollars. 
Find  the  sum  and  the  rate  of  interest. 

46.  A  person  has  a  certain  capital  invested  at  a  certain  rate 

per  cent.  Another  person  has  %  1000  more  capital, 
and  his  capital  invested  at  one  per  cent  better  than 
the  first,  and  receives  an  income  %  80  greater.  A  third 
person  has  $1500  more  capital,  and  his  capital  in- 
vested at  two  per  cent  better  than  the  first,  and. re- 
ceives an  income  %  150  greater.  Find  the  capital  of 
each,  and  the  rate  at  which  it  is  invested. 

47.  A  person  has  $12,750  to  invest.     He  can  buy  three  per 

cent  bonds  at  81,  and  five  per  cents  at  120.  Find 
the  amount  of  money  he  must  invest  in  each  in  order 
to  have  the  same  income  from  each  investment. 

48.  A  and  B  each  invested  $  1500  in  bonds ;    A  in  three 

per  cents  and  B  in  four  per  cents.  The  bonds  were 
bought  at  such  prices  that  B  received  $5  interest 
more  than  A.  After  both  classes  of  bonds  rose  10 
points,  they  sold  out,  and  A  received  $50  more  than 
B.     What  price  was  paid  for  each  class  of  bonds? 


PROBLEMS.  177 


49.  A  person  invests  $10,000  in  three  per  cent  bonds, 
$16,500  in  three  and  one-half  per  cents,  and  has  an 
income  from  both  investments  of  $1056.25.  If  his 
investments  had  been  $2750  more  in  the  three  per 
cents,  and  less  in  the  three  and  one-half  per  cents, 
his  income  v/ould  have  been  62|-  cents  greater. 
What  price  was  paid  for  each  class  of  bonds  ? 

50o  The  sum  of  $2500  was  divided  into  two  unequal  parts 
and  invested,  the  smaller  part  at  two  per  cent  more 
than  the  larger.  The  orde  of  interest  on  the  larger 
sum  was  afterwards  increased  by  1,  and  that  of  the 
smaller  sum  diminished  by  1 ;  and  thus  the  interest 
of  the  whole  was  increased  by  one-fourth  of  its  value. 
If  the  interest  of  the  larger  sum  had  been  so  in- 
creased, and  no  change  been  made  in  the  interest  of 
the  smaller  sum,  the  interest  of  the  whole  would  have 
been  increased  one-third  of  its  value.  Find  the  sums 
invested,  and  the  rate  per  cent  of  each. 

Note  VII.  If  x  represent  the  number  of  linear  units  in  the  length, 
and  y  in  the  width,  of  a  rectangle,  xy  will  represent  the  number  of 
its  units  of  surface ;  the  surface  unit  having  the  same  name  as  the 
Hnear  unit  of  its  sides. 

51.  If  the  sides  of  a  rectangular  field  were  each  increased 

by  2  yards,  the  area  would  be  increased  by  220 
square  yards ;  if  the  length  were  increased  and  the 
breadth  were  diminished  each  by  5  yards,  the  area 
would  be  diminished  by  185  square  yards.  What  is 
its  area  ? 

52.  If  a  given  rectangular  floor  had  been  3  feet  longer  and 

2  feet  broader  it  would  have  contained  64  square  feet 
more ;  but  if  it  had  been  2  feet  longer  and  3  feet 
broader  it  would  have  contained  68  square  feet  more. 
Find  the  length  and  breadth  of  the  floor. 


178  ALGEBRA. 


53.  In  a  certain  rectangular  garden  there  is  a  strawberry- 

bed  wbose  sides  are  one-third  of  the  lengths  of  the 
corresponding  sides  of  the  garden.  The  perimeter  of 
the  garden  exceeds  that  of  the  bed  by  200  yards ; 
and  if  the  greater  side  of  the  garden  be  increased  by 
3,  and  the  other  by  5  yards,  the  garden  will  be  en- 
larged by  645  square  yards.  Find  the  length  and 
breadth  of  the  garden. 

Note  VIII.   Care  must  be  taken  to  express  the  conditions  of  a 
problem  with  reference  to  the  same  principal  unit. 

Ex.  In  a  mile  race  A  gives  B  a  start  of  20  yards  and  beats 
him  by  30  seconds.  At  the  second  trial  A  gives  B  a 
start  of  32  seconds  and  beats  him  by  9^  yards. 
Find  the  rate  per  hour  at  which  each  runs. 

Let  X  =  number  of  yards  A  runs  a  second, 
and     y  =  number  of  3'ards  B  runs  a  second. 
Since  there  are  1760  yards  in  a  mile, 

.  =  number  of  seconds  it  takes  A  to  run  a 

^  mile, 

and  — — ~  =  number  of  seconds  B  was  running  in  the 

^  ^  first  and  second  trials,  respectively. 

Hence.  1!!5_1^  =  30. 
and       W_l^=32 

y         ^ 

The  solution  of  these  equations  gives  x  =  5^|  and  y  =  5^j. 

That  is,  A  runs  -^^,  or  ,  of  a  mile  in  one  second  ; 

1760         300 
and  in  one  hour,  or  3600  seconds,  runs  12  miles. 
Likewise,  B  runs  lOj^/j  miles  in  one  hour. 

54.  In  a  mile  race  A  gives  B  a  start  of  100  yards  and  beats 

him  by  15  seconds.  In  the  second  trial  A  gives  B  a 
start  of  45  seconds  and  is  beaten  by  22  yards.  Find 
the  rate  of  each  in  miles  per  hour. 


PROBLEMS.  179 


55.  In  a  mile  race  A  gives  B  a  start  of  44  yards  and  beats 

him  by  51  seconds.     In  the  second  trial  A  gives  B  a 
.     start  of  1  minute  and  15  seconds  and  is  beaten  by  88 
yards.     Find  the  rate  of  each  in  miles  per  hour. 

56.  The  time  which  an  express-train  takes  to  go  120  miles 

is  Y^^  of  the  time  taken  by  an  accommodation-train. 
The  slower  train  loses  as  much  time  in  stopping  at 
different  stations  as  it  would  take  to  travel  20  miles 
without  stopping ;  the  express-train  loses  only  half 
as  much  time  by  stopping  as  the  accommodation- 
train,  and  travels  15  miles  an  hour  faster.  Find  the 
rate  of  each  train  in  miles  per  hour. 

67.  A  train  moves  from  P  towards  Q,  and  an  hour  later  a 
second  train  starts  from  Q  and  moves  towards  P 
at  a  rate  of  10  miles  an  hour  more  than  the  first 
train ;  the  trains  meet  half-way  between  P  and  Q. 
If  the  train  from  P  had  started  an  hour  after  the 
train  from  Q  its  rate  must  have  been  increased  by  28 
miles  in  order  that  the  trains  should  meet  at  the 
half-way  point.     Find  the  distance  from  P  to  Q. 

58.  A  passenger-train,  after  travelling  an  hour,  meets  with 

an  accident  which  detains  it  one-half  an  hour ;  after 
which  it  proceeds  at  four-fifths  of  its  usual  rate,  and 
arrives  an  hour  and  a  quarter  late.  If  the  accident 
had  happened  30  miles  farther  on,  the  train  would 
have  been  only  an  hour  late.  Determine  the  usual 
rate  of  the  train. 

59,  A  passenger-train  after  travelling  an  hour  is  detained 

15  minutes ;  after  which  it  proceeds  at  three-fourths 
of  its  former  rate,  and  arrives  24  minutes  late.  If 
the  detention  had  taken  place  5  miles  farther  on,  the 
train  would  have  been  only  21  minutes  late.  Deter- 
mine the  usual  rate  of  the  train. 


180  ALGEBRA. 


60.  A  man  bought  10  oxen,  120  sheep,  and  46  lambs.     The 

cost  of  3  sheep  was  equal  to  that  of  5  lambs ;  an  ox, 
a  sheep,  and  a  lamb  together  cost  a  number  of  dol- 
lars less  by  57  than  the  whole  number  of  animals 
bought;  and  the  whole  sum  spent  was  §2341.50. 
Find  the  price  of  an  ox,  a  sheep,  and  a  lamb,  respec- 
tively. 

61.  A  farmer  sold  100  head  of  stock,  consisting  of  horses, 

oxen,  and  sheep,  so  that  the  whole  realized  §11.75  a 
head ;  while  a  horse,  an  ox,  and  a  sheep  w^ere  sold 
for  $110,  $62.50,  and  $7.50,  respectively.  Had  he 
sold  one-fourth  of  the  number  of  oxen  that  he  did, 
and  25  more  sheep,  he  would  have  received  the  same 
sum.  Find  the  number  of  horses,  oxen,  and  sheep, 
respectively,  which  were  sold. 

62.  A,  B,  and  C  together  subscribed  $  100.     If  A's  sub- 

scription had  been  one-tenth  less,  and  B's  one-tenth 
more,  O's  must  have  been  increased  by  $2  to  make 
up  the  sum ;  but  if  A's  had  been  one-eighth  more, 
and  B's  one-eighth  less,  C's  subscription  would  have 
been  $  17.50.     What  did  each  subscribe  ? 

63.  A  gives  to  B  and  C  as  much  as  each  of  them  has ;  B 

gives  to  A  and  0  as  much  as  each  of  them  then  has ; 
and  C  gives  to  A  and  B  as  much  as  each  of  them 
then  has.  In  the  end  each  of  them  has  $6.  How 
much  had  each  at  first  ? 

64.  A  pays  to  B  and  0  as  much  as  each  of  them  has ;  B 

pays  to  A  and  0  one-half  as  much  as  each  of  them  then 
has ;  and  G  pays  to  A  and  B  one-third  of  what  each  of 
them  then  has.  In  the  end  A  finds  that  he  has 
$1.50,  B  $4.16|,  0  $.581  How  much  had  each  at 
first? 


CHAPTER  XIII. 

Involution  and  Evolution. 

198.  The  operation  of  raising  an  expression  to  any  re- 
quired power  is  called  Involution. 

Every  case  of  involution  is  merely  an  example  of  multi- 
plication, in  which  the  factors  are  equal.     Thus, 
{2ay  =  '2,a^x2a^  =  ^a\ 

199.  A  power  of  a  simple  expression  is  found  by  multi- 
plying the  exponent  of  each  factor  by  the  exponent  of  the 
required  factor,  and  taking  the  product  of  the  resulting 
factors.  The  proof  of  the  law  of  exponents,  in  its  general 
form,  is :       (^«j»  =  «-  x  a"»  X  a*"  X to  n  factors, 

~m  +  m  +  w  + to  n  terma, 

Hence,  if  the  exponent  of  the  required  power  be  a  com- 
posite number,  it  may  be  resolved  into  prime  factors,  the 
power  denoted  by  one  of  these  factors  may  be  found,  and 
the  result  raised  to  a  power  denoted  by  another,  and  so  on. 
Thus,  the  fourth  power  may  be  obtained  by  taking  the  sec- 
ond power  of  the  second  power ;  the  sixth  by  taking  the 
second  power  of  the  third  power ;  the  eighth  by  taking  the 
second  power  of  the  second  power  of  the  second  power. 

200.  From  the  Law  of  Signs  in  multiplication  it  is  evi- 
dent that, 

I.    All  even  powers  of  a  number  are  positive. 
II.    Ali  odd  powers  of  a  number  have  the  same  sign  as 
the  number  itself. 


182  ALGEBRA. 


Hence,  no  even  power  of  any  number  can  be  negative; 
and  of  two  compound  expressions  whose  terms  are  identical 
but  have  opposite  signs,  the  even  powers  are  the  same. 
Thus, 

{b  -af  =  \-{a--b)\^  ^  {a-bf. 

201.  A  method  has  been  given,  §  83,  of  finding,  without 
actual  multiplication,  the  powers  of  binomials  which  have 
the  form  (a  dz  ^). 

The  same  method  may  be  employed  when  the  terms  of  a 
binomial  have  coeficients  or  exponents. 

(2)  {px^-lff, 

=  {bx^f  -  3  {bx^)\2,f)  +  Z(bx^  (2ff  ~  (2if)\ 

=  i25x«  - 160  o^y + mxY  -  %if. 

(3)  {a~by=-a^-4:a^b-\-QaW-^ab^  +  b\ 

(4)  {x'-lyy, 

=^x^~  2xhj  +  ixV  -  ^xY  +  i^y\ 

202.  In  like  manner,  a  'polynomial  of  three  or  more  terms 
may  be  raised  to  any  power  by  enclosing  its  terms  in  par- 
entheses, so  as  to  give  the  expression  the  form  of  a  binomial. 
Thus, 

(1)    {a-\-b^cf=\a-\-{}>-^e)\^ 

=  a^  +  3a2  (b^c)  +  Za{h-\-  c)2+  (b  +  cf, 
=  a^  +  Sa^b  +  3a^c  +  Sab^  +  Qabc 

+  3ac-2  +  b^ -\-Sb'c  \-6b(^  +  c\ 


INVOLUTION    AND    EVOLUTION.  183 

C2)    (a^-2a^  +  Sx  +  4:y, 

=  x^-4:Cc^+4:x'+Qx*-4:x^-16x'+da^+24:X-{-16, 
=  ^6 _  4^  _!_  1Q^,4  _  4^  __  7^  _|,  24a;  +  16. 

Exercise  LXXVI. 
Write  the  second  members  of  the  following  equations : 


1.    (a?f  = 

11.    (2a26c3)^  = 

21.    (-3a'b'cy  = 

2.    (xy  = 

12.    (-5a:r«y2^)«  = 

22.    (-3^/  = 

3.    (^/)2  = 

13.    (-7m^nx^/y  = 

=  23.    (-5a'bj^y  = 

'■  (f  T= 

V     SabcJ 

-  (-ST= 

'  m- 

15.    (3^+1/  = 

» e^^j- 

a  (^  +  2/  = 

16.    {2x-ay--= 

26.    (l-a~-ay^ 

7.    (x-2)*  = 

17.    (3:?;  +  2a/  = 

27.    (2-3:r+4:r2)3_ 

8.    (a;  +  37- 

18.    (2:r-y)^  = 

28.    (l-2x+ar'f  = 

9.    (l  +  2:r7-- 

19.    (x'i/-2x2/y  = 

29.    (l-:r+:r2)3=3 

LO.    {2m-iy  = 

20.    (aZ^-3y  = 

30.    (l+:ir+r^)*  = 

Evolution. 

203.  The  operation  of  finding  any  required  root  of  an 
expression  is  called  Evolution. 

Every  case  of  evolution  is  merely  an  example  of  factor- 
ing, in  which  the  required  factors  are  all  equal.     Thus, 

the  square,  cube,  fourth roots  of  an  expression  are  found 

by  taking  one  of  the  two,  three,  four equal  factors  of  the 

expression. 


184  ALGEBRA. 

204.  The  symbol  which  denotes  that  a  square  root  is  to 
be  extracted  is  -y/ ;  and  for  other  roots  the  same  symbol  is 
used,  but  with  a  figure  written  above  to  indicate  the  root, 
thus,  -^,  -J/,  etc.,  signifies  the  third  root,  fourth  root,  etc. 

205.  Since  the  cube  of  a^  =  a^,  the  cube  root  of  a^  —  a^. 

Since  the  fourth  power  of  2a^  =  2^a^,  the  fourth  root  of 
2*a«  =  2a2. 

Since  the  square  of  abc  =  ct^b^(^,  the  square  root  of  a^b^c^ 
^=abc. 

o-        1.x.  coh      c^b^  ,T  .    r;C?b^      ab 

bince  the  square  oi  —  =  — — ,  the  square  root  oi  -^^.  =  — 

xy     oT'if  orif      xy 

Hence,  the  root  of  a  simple  expression  is  found  by  divid- 
ing the  exponent  of  each  factor  by  the  index  of  the  root,  and 
talcing  the  product  of  the  resulting  factors. 

206.  It  is  evident  from  §  200  that 

I.   Any  even  root  of  a  positive  number  will  have  the 
double  sign,  ±. 

II.    There  can  be  no  even  root  of  a  negative  number. 

III.  Any  odd  root  of  a  number  will  have  the  same  sign  as 
the  number. 


4 


81  a^  3a*  * 


But  V— ?  is  neither  -|-  x  nor  —  x,  for  (-f  x)^  =  -j-  x^,  and 
(-x/  =  -i-a^. 

The  indicated  even  root  of  a  negative  number  is  called 
an  impossible,  or  imaginary,  number. 


INVOLUTION  AND  EVOLUTION. 


185 


207.  If  the  root  of  a  number  expressed  in  figures  is  not 
readily  detected,  it  may  be  found  by  resolving  the  number 
into  its  prime  factors.  Thus,  to  find  the  square  root  of 
3,415,104 : 


2« 

3415104 

2' 

426888 

S' 

53361 

7 

5929 

7 

847 

11 

121 

11 

.-.      3,415,104  =  2«  X  32  X  72  X  IP. 

.-.  V3,415,104  =  2^x3  x7  xll  =  1848. 


o-      Tf  Exercise  LXXVII. 

bimpliiy  : 

1.    V^^  a/^,  V4^2^  </M,  ■y/a'x'y',  </l6^^W?,  ^f^Ma""". 


2.   V-1728c'«(^V/,    \/3375PV^  V3111696ciV. 

3.  V5336uvyv«,  jIZME^,  ^m^, 

^  \       3432^^'     \72923o 

4.  V25  a^Z^V  +  a/8^s^V  -  ■</81  a'h^c*  -  -y/W^^^. 

5.  ^27^  X  -v/MsT?  X  Vl6^. 

When  a  =  l,  ^  =  3,  rr  =  2,  y  =  6,  find  the  values  of ; 

6.  4 V2x  —  Vahxy  4-  5 ^c^h^xy. 

7.  2a-yJ^x-\'hi/\2hy^^ahx^hxy. 

8.  V«2 +"2^5T'^2  ><  ^of^Za%-^Zah''-\-h^. 

9.  -v/^>«^^3l2a  +  3^a2-a3 -  V^' +^2^^^^2^^>. 


186  ALGEBRA. 


Square  Eoots  of  Compound  Expressions. 

208.  Since  the  square  oi  a-\-b  is  a'^^2ab-{-  h"^,  the  square 
root  of  a^  -f-  2 (2^  +  5^  is  a-{-h. 

It  is  required  to  find  a  method  of  extracting  the  root 
a-\-h  when  a^-{-2ab  -\-h^  is  given  : 

Ex.  The  first  term,  a,  of  the  root  is  obviously  the  square  root  of  the 
first  term,  a',  in  the  expression. 

a^  -{-  2ah  -\-  h'^\a  -{-  h  ^^  ^^®  ^  ^®  subtracted  from  the 

2  given  expression,  the  remainder  is 

o       I   tT"      o    7,    I    72  2ah->rW.   Therefore  the  second  term, 

Za+b        Zab  +  b^  ^^  ^^  ^^^  ^^^^  -^  obtained  when  the 

Zap  -\-  0'  £j.g{;  iQj-Ya  of  this  remainder  is  di- 

vided by  2  a,  that  is,  by  double  the 
part  of  the  root  already  found.  Also,  since  2ab  +  b'^  =  (2a  +  h)b,  the  di- 
visor is  completed  by  adding  to  the  trial-divisor  the  new  term  of  the  root. 

(1)    Find  the  square  root  of  25:^;^  —  20a^ij  -f  4a;y. 
-20:r3y  +  4a:y 


10a7-2a;2^ 


The  expression  is  arranged  according  to  the  ascending  powers  of  x. 

The  square  root  of  the  first  term  is  bx,  and  5  a;  is  placed  at  the 
right  of  the  given  expression,  for  the  first  term  of  the  root. 

The  second  term  of  the  root,  —  2a^?/,  is  obtained  by  dividing 
—  20o?y  by  10a;,  and  this  new  term  of  the  root  is  also  annexed  to  the 
divisor,  10a;,  to  complete  the  divisor. 

209.  The  same  method  will  apply  to  longer  expressions, 
if  care  be  taken  to  obtain  the  trial-divisor  at  each  stage  of 
the  process,  by  doubling  the  part  of  the  foot  already  found, 
and  to  obtain  the  complete  divisor  by  annexing  the  new  term 
of  the  root  to  the  trial-divisor. 


INVOLUTION   AND    EVOLUTION.  187 

Ex.    Find  the  square  root  of 

16  x^~24:  .1^+25  2:^-20  :?:3+10  ^-4  x+1  \4:X^-3a^+2x-l 


8^-3a;2 


~24.a^-}-2bx* 
-24^^+  9x* 


8a^-(ja^-i-2x 


16:r^-20a;3+10ar' 
16r?;*-12a:3_|_  4^ 


8:r3-6:r2+4:r~l 


-  8x^+  6:^-4a;+l 

-  8:^+  6x'-4:x+l 


The  expression  is  arranged  according  to  the  descending  powers 
of  X. 

It  will  be  noticed  that  each  successive  trial- divisor  may  be  obtained 
by  taking  the  preceding  complete  divisor  with  its  last  term  doubled. 

Exercise  LXXVIII. 
Extract  the  square  roots  of : 

1.  a^  +  4a3  +  2a2-4a  +  l. 

2.  x^~2a^i/-]-Sx'f-2x7/  +  7/. 

3.  4r/-12a^.^^  +  5aV+6aV  +  A*. 

4.  9x^-12c(^/+l6a^i/*-24:xy-{-4:y'^-{-16x7/. 

5.  4t6»+166-«+16aV-32aV. 

6.  4:x*+9~30x-20x^  +  37:i^. 

7.  IQx* ~  IGahx"  +  l^h^x"  +  Aa^b^ -  8ah^  +  U\ 

8.  x^  +  25x'+l0x^-ix^-^20x^  +  16-24:x. 

9.  x^ +  8xy~Ax^i/- 4x1/ -\~8xy- 100^2/ -{-y\ 

10.  4-12a-lla'^  +  5a2-4a^  +  4a«+14a^ 

11.  da^--Gab  +  30ac-{'6adi-b^-10hc-2bd 

+  256-2+ 10ccZ+J2. 


188  ALGEBRA. 


12.  25a;«  -  31  a^y  +  34  :p3y3  -  SOor'y  +  y«  _  8^/  +  10^;^. 

13.  m^  -  4m^  +  10  m^  -  20  m^  -  44  m^ 

+  35mH46m2-40m  +  25. 

14.  x^  —  3?y  —  -oi?y^-\-x'i^-\-\f. 

15.  ;r''-4:c33/  +  6:ry-6:iy  +  5y^-^  +  ^. 

a;       ar 

16.  ^'_-^  +  43^2^^_3^^,3^ 
9        2       48  4  4 


X         Ol? 


x^ 


b^       b  a       o?  ^         12      3      9 


Square  Roots  of  Arithmetical  Numbers. 

210.  In  the  general  method  of  extracting  the  square  root 
of  a  number  expressed  by  figures,  the  first  step  is  to  mark 
ofi*  the  figures  in  'periods. 

Since  1  =  1^,  100  =  10^  10,000  =  100=,  and  so  on,  it  is  evident  that 
the  square  root  of  any  number  between  1  and  100  hes  between  1  and 
10  ;  the  square  root  of  any  number  between  100  and  10,000  Hes  be- 
tween 10  and  100.  In  other  words,  the  square  root  of  any  number 
expressed  by  one  or  two  figures  is  a  number  of  one  figure ;  the  square 
root  of  any  number  expressed  by  three  or  Jour  figures  is  a  number  of 
two  figures  ;  and  so  on. 

If,  therefore,  a  dot  be  placed  over  the  units  figure  of  a  square  num- 
ber, and  over  every  alternate  figure,  the  number  of  dots  will  be  equal 
to  the  number  of  figures  in  its  square  root. 

Find  the  square  root  of  3249. 

3249  (57  In  this  case,  a  in  the  typical  form  a^  +  2a5  -f-  6* 

25  represents  5  tens,  that  is,  50,  and  h  represents  7. 

107)  749  The  25  subtracted  is  really  2500,  that  is,  a^,  and 

749  the  complete  divisor,  2a  +  6,  is  2  X  50  +  7  =  107. 


INVOLUTION   AND   EVOLUTION.  189 

211.  The  same  metHod  ,will  apply  to  numbers  of  more 
than  two  periods  by  considering  a  in  the  typical  form  to 
represent  at  each  step  the  part  of  the  root  already  found. 

It  must  be  observed  that  a  represents  so  many  tens  with  respect  to 
the  next  figure  of  the  root. 

Ex.    Find  the  square  root  of  5,322,249. 

5322249(2307 

4        ' 
43)132 

129 

4607)32249 
32249 

212.  If  the  square  root  of  a  number  have  decimal  places, 
the  number  itself  will  have  twice  as  many. 

Thus,  if  .21  be  the  square  root  of  some  number,  this  number  will 
be  (.21)'  =  .21  X  .21  =  .0441 ;  and  if  .111  be  the  root,  the  number  will 
be  (.111)2  _  111  X  111  _  .012321. 

Therefore,  the  number  of  decimal  places  in  every  square  decimal 
will  be  even,  and  the  number  of  decimal  places  in  the  root  will  be 
half  as  many  as  in  the  given  number  itself 

Hence,  if  the  given  square  number  contain  a  decimal,  and  a  dot  be 
placed  over  the  units^  figure,  and  then  over  every  alternate  figure  on 
both  sides  of  it,  the  number  of  dots  to  the  left  of  the  decimal  point 
will  show  the  number  of  integral  places  in  the  root,  and  the  number 
of  dots  to  the  right  will  show  the  number  of  decimal  places. 

Ex.    Find  the  square  roots  of  41.2164  and  965.9664. 

4l.2i64  (  6.42  965.9664  (  31.08 

36  9_ 

124)521  61)65 

496  61 


1282)2564  6208)49664 

2564  49664       • 

It  is  seen  from  the  dotting  that  the  root  of  the  first  example  will 
have  one  integral  and  two  decimal  places,  and  that  the  root  of  the 
second  example  will  have  two  integral  and  two  decimal  places. 


190  ALGEBRA. 


213.  If  a  number  contain  aji  odd  number  of  decimal 
places,  or  if  any  number  give  a  remainder  when  as  many 
figures  in  the  root  have  been  obtained  as  the  given  number 
has  periods,  then  its  exact  square  root  cannot  be  found. 
We  may,  however,  approximate  to  its  exact  root  as  near  as 
we  please  by  annexing  ciphers  and  continuing  the  operation. 

Ex.    Find  the  square  roots  of  3  and  357.357. 


3.(1.732 

357.3570(18.903 

1 

27)200 
189 

28)257 
224 

343)1100 
1029 

369)3335 
3321 

3462)7100 
6924 

37803)147000 
113409 

Exercise  LXXIX. 
Extract  the  square  roots  of: 

1.  120,409;  4816.36;  1867.1041;  1435.6521;  64.128064. 

2.  16,803.9369;  4.54499761;  .24373969;  .5687573056. 

3.  .9;  6.21;  .43;  .00852;  17;  129;  347.259. 

4.  14,295.387;  2.5;  2000;   .3;   .03;  111. 

5.  .00111;  .004;  .005;  2;  5;  3.25;  8.6. 

a       1.16.     100.     169  .     289  .     400 
"•     ¥'    ¥9'    TT4'    YT5  ^    3T¥  >    W2Z- 

7       1.2.3.      1     .        7      .        6.6.1 

'•    2" '   3" '  4  '  t2  '  tts  '  t2t  »  t  >  t2- 

Cube  Roots  of  Compound  Expressions. 

214.  Since  the  cube  of  a  -f  ^  is  a^  +  3  a^b  +  3  ab^  +  ^^  the 
cube  root  of  a^  +  3  a^b  +  3  ai^  +  Z>^  is  a  +  b. 

It  is  required  to  find  a  method  for  extracting  the  cube 
root  a-\-b  when  a^  +  3  a^b  -{-  3  ab^  -\-  ¥  is  given : 


INVOLUTION    AND    EVOLUTION.  191 

(1)    Find  the  cube  root  of  a^  +  3  a%  -\-Zah'-\-  h^, 
+  3a5  +  ^'     3a25  +  3a62+J3 


3a2+3a6  +  Z>2     Za% -\-Za}y'^V' 


The  first  term  a  of  the  root  is  obviously  the  cube  root  of  the  first 
term  a^  of  the  given  expression. 

If  a^  be  subtracted,  the  remainder  is  3a^6  +  3a6^  +  6^;  therefore, 
the  second  term  h  of  the  root  is  obtained  by  dividing  tJie  first  term 
of  this  remainder  by  three  times  the  square  of  a. 

Also,  since  3a^b  +  3  ah^  +  &^  =  (3  a*^  +  3  a6  +  h^)  h,  the  complete  divisor 
is  obtained  by  adding  3 ah  +  h^  to  the  trial-divisor  3a^. 

(2)    Find  the  cube  root  oi^s^ +  2>e>x^y -{-  54  ^  +  21  f. 

8  a;8-f  36  a?V+54  xf-{-21f  '2x+?>y 
12.t2  8^ 


(6a;+3y)  3y=  18a:?/+9: 


12a;2_|_i8^^_|_9^ 


36a;2y+54:ry2-f27/ 
363;'y-f54a;?/4-27?/ 


The  cube  root  of  the  first  term  is  2  a;,  and  this  is  therefore  the  first 
term  of  the  root. 

The  second  term  of  the  root,  3  y,  is  obtained  by  dividing  36  x^y  by 
3{2xf=  12a;^,  which  corresponds  to  3a'  in  the  typical  form,  and  is 
completed  by  annexing  to  1 2 a;^  the  expression  |3(2a;)  +  37/|32/  =  18a;2/ 
+  92/*,  which  corresponds  to  3a6  +  5*,  in  the  typical  form. 

215.  The  same  method  may  be  applied  to  longer  expres- 
sions by  considering  a  in  the  typical  form  3  a^  -f  3  a5  +  Z>^  to 
represent  at  each  stage  of  the  process  the  part  of  the  root 
already  found. 

Thus,  if  the  part  of  the  root  already  found  be  a;  +  y,  then  3  a^  of 
the  typical  form  will  be  represented  by  3  (a;  +  yf ;  and  if  the  third 
term  of  the  root  be  +  z,  the  3  a6  +  6'  will  be  represented  by  3  (a;  +  y)  a 
+  2*.  So  that  the  complete  divisor,  3  a*  +  3  a6  -»-  b^,  will  be  repre- 
sented by  3  (a;  +  2/)'  +  3  (a;  +  y)  2  +  2*. 


192  ALGEBRA. 


Find  the  cube  root  of  .i'^  -  3:r^  +  5a;^  -  3^;  —  1. 
(3a^-a;)(- 


3(x2-ir)2  = 
(3x«_3a;-l)(-l)  = 


3x*-3x3+    a^|-3a;5 

+  3ir4-    ar* 

30.-* -6^-3  + Bar' 

-3a;2  +  3a;  +  l 

-3a;4  +  6a:3„3  3,_i 

3a^-6a;3            +3a;  +  l 

-3x4  +  6a^-3a;-l 

The  root  is  placed  above  the  given  expression  for  convenience  of 
arrangement. 

The  first  term  of  the  root,  o?,  is  obtained  by  taking  the  cube  root 
of  the  first  term  of  the  given  expression ;  and  the  first  trial-divisor, 
3ar*,  is  obtained  by  taking  three  times  the  square  of  this  term  of  the 
root. 

The  first  complete  divisor  is  found  by  annexing  to  the  trial-divisor 
(3  x'^  —  x)  (—  a;),  which  expression  corresponds  to  (3  a  -h  6)  6  in  the 
typical  form. 

The  part  of  the  root  already  found  (a)  is  now  represented  by  a;^  —  a; ; 
therefore  3  a*  is  represented  by  3  (a;^  —  a)^  =  3  a^  —  6  a;^  +  3  x^,  the  sec- 
ond trial-divisor ;  and  (3  a  +  &)  &  by  (3a;2  —  3  a;  —  1)  (—  1) ;  therefore, 
in  the  second  complete  divisor,  3  a''  -f  (3  a  +  &)  5  is  represented  by 

{^x^-Q^  +  ?,o?)  +  {-2,x'^-Zx-V)x{~l)  =  2>x^-Q>x^-\-Zx  +  l. 

Exercise  LXXX. 
Find  the  cube  roots  of : 

1.  37^+6:^2^+12^+82/^.       3.  :i;«  +  125;2  +  48rc  +  64. 

2.  a3-9a2  +  27a-27.  4.  x^-Zaa^+ba?j^-Za^x-a\ 

5.  :r«  +  3:r^  +  65;^+7:i;3+6a,^  +  3:i7+l. 

6.  l-9:r  +  39:^-99:r3+1565;^-144.r^  +  64a;«. 

7.  a«-6a*  +  9a^  +  4a«-9a2-6a-l. 

8.  64:r«  +  1920*^  +  144:^^^  -  32:^^  -  36:r2+  12a:  -  1. 

9.  l-^x-\-^x^-lOa^+l2x^-\^o^-\-\Ox^-Q>x'+Zx^~x\ 


INVOLUTION   AND   EVOLUTION.  193 

10.  a«  +  9  a'b  -  135  a%^  +  729  ah'  -  729  b\ 

11.  c"  - 12  b(^  +  60  5V  - 160  6V  +  240  ^»V  - 192  h'c  +  64  Z/«. 

12.  3  a«  +  48  a^5  +  60  a^b""  -  80  a^^^  -  90  a'^b^i- 108  aZ)^-  27^^ 

Cube  Roots  of  Arithmetical  Numbers. 

216.  In  extracting  the  cube  root  of  a  number  expressed 
by  figures,  the  first  step  is  to  mark  it  off  into  periods. 

Since  1  =  l^,  1000=  lO^,  1,000,000  =  lOO^,  and  so  on,  it  follows  that 
the  cube  root  of  any  number  between  1  and  1000,  that  is,  of  any  num- 
ber which  has  one,  two,  or  three  figures,  is  a  number  of  one  figure; 
and  that  the  cube  root  of  any  number  between  1000  and  1,000,000, 
that  is,  of  any  number  which  h.a,a  four.  Jive,  or  six  figures,  is  a  number 
of  two  figures  ;  and  so  on. 

Hence,  if  a  dot  be  placed  over  every  tJiird  figure  of  a  cube  num- 
ber, beginning  with  the  units'  figure,  the  number  of  dots  will  be  equal 
to  the  number  of  figures  in  its  cube  root. 

217.  If  the  cube  root  of  a  number  contain  any  decimal 
figures,  the  number  itself  will  contain  three  times  as  many. 

Thus,  if  .3  be  the  cube  root  of  a  number,  the  number  is  .3  X  .3  X  .3 
-=  .027. 

Hence,  if  the  given  cube  number  have  decimal  places,  and  a  dot 
be  placed  over  the  units'  figure  and  over  every  third  figure  on  both 
sides  of  it,  the  number  of  dots  to  the  left  of  the  decimal  point  will 
show  the  number  of  integral  figures  in  the  root ;  and  the  number  of 
dots  to  the  right  will  show  the  number  of  decimal  figures  in  the  root. 

If  the  given  number  be  not  a  perfect  cube,  ciphers  may  be  an- 
nexed, and  a  value  of  the  root  may  be  found  as  near  to  the  true  value 
as  we  please. 

218.  It  is  to  be  observed  that  if  a  denote  the  first  term 
of  the  root,  and  b  the  second  term,  the  first  comjplete  divisor 

^^  3a2  +  3ai  +  Z.2, 

and  the  second  trial-divisor  is  3  (a  -j-  hy,  that  is, 

3a2  +  6a^  +  352, 


194 


ALGEBRA. 


which  may  be  obtained  from  the  preceding  complete  divisor 
\>y  adding  to  it  its  second  term  and  twice  its  third  term, 

a  method  which  will  very  much  shorten  the  work  in  long 
arithmetical  examples. 

219.    Ex.    Extract  the  cube  root  of  5  to  five  places  of 
decimals. 


5.000(1.70997 
1 


3  X  10^  =  300 

8(10x7)  =  210 

T=-    49 

559 

259 

3  X  1700^ 


4  000 


3  913 


8670000 
3(1700x9)=     45900 
9'^= 8^] 

8715981 
45981 


5  x  1709^  =  8762043 


87  000  000 


443  829 


8  556  1710 
7  885  8387 


670  33230 
613  34301 


After  the  first  two  figures  of  the  root  are  found,  the  next  trial  divi- 
sor is  obtained  by  bringing  down  the  sum  of  the  210  and  49  obtained 
in  completing  the  preceding  divisor ;  then  adding  the  three  lines  con- 
nected by  the  brace,  and  annexing  two  ciphers  to  the  result. 

The  last  two  figures  of  the  root  are  found  by  division.  The  rule  in 
such  cases  is,  that  two  less  than  the  number  of  figures  already  ob- 
tained may  be  found  without  error  by  division,  the  divisor  to  be  em- 
ployed being  three  times  the  square  of  the  part  of  the  root  already 
found. 


involution  and  evolution.  195 

Exercise  LXXXI. 
Find  the  cube  roots  of : 

1.  274,625.               7.    1601.G13.  13.  33,076.161. 

2.  110,592.               8.    1,259,712.  14.  102,503.232. 

3.  262,144.               9.   2.803221.  15.  820.025856. 

4.  884.736.             10.    7,077,888.  16.  8653.002877. 

5.  109,215,352.       11.    12.812904.  17.  1.371330631. 

6.  1,481,544.          12.   56.623104.  18.  20,910.518875. 
19.   91.398648466125.            20.  5.340104393239. 

21.    Find  to  four  figures  the  cube  roots  of  2.5  ;  .2 ;  .01 ;  4 ;  .4. 

220.  Since  the  fourth  power  is  the  square  of  the  square, 
and  the  sixth  power  the  square  of  the  cube ;  the  fourth  root 
is  the  square  root  of  the  square  root,  and  the  sixth  root  is 
the  cube  root  of  the  square  root.  In  like  manner,  the 
eighth,  ninth,  twelfth roots  may  be  found. 

Exercise  LXXXII. 
Find  the  fourth  roots  of: 

1.  81  a'  -  540  a^b  +  1350  a^'b^  -  1500  ab^  +  625  b\ 

2.  1  -  4:r  +  lOrr^  -  16a;3_^  19^4  _jL6^-flO:c«-4a;^  + 0:8. 

Find  the  sixth  roots  of : 

3.  64  -  192.^  +  240^:2  _.  iqq,^:^  ^  gOo;^  -  12:c^  +  x\ 

4.  729 a:« -  1458^;^  +  1215a;*  - 540^;='  +I2>b:^ -l^x+l. 

Find  the  eighth  root  of : 

5.  1  -  8y  +  28?/2  _  55^  _^  79 yi ._  55^!  _|_  28/  -  8y^  +  f. 


CHAPTER  XIV. 

Quadratic  Equations. 

221.  An  equation  wHcli  contains  the  square  of  the  un- 
known quantity,  but  no  higher  power,  is  called  a  quadratic 
equation. 

222.  If  the  equation  contain  the  square  only,  it  is  called 
a  pure  quadratic ;  but  if  it  contain  the  first  power  also,  it  is 
called  an  affected  quadratic. 

Pure  Quadratic  Equations. 

Solve  the  equation  5:r^  —  48  =  2:r^. 

5^  —  43  =  2^:^  It  will  be  observed  that  there  are  two  roots  of 

3  ^  _.  ^g         equal  value  but  of  opposite  signs  ;  and  there  are 

™2  __  1  p         only  two,  for  if  the  square  root  of  the  equation, 

,    .       rc^  =  16,  were  written  ±  a;  =  ±  4,  there  would  be 

only  two  values  of  x ;    since  the  equation  —  x 

=  +  4  gives  .T  =  —  4,  and  the  equation  —  a;  =  —  4  gives  re  =  4. 

Hence,  to  solve  a  pure  quadratic. 

Collect  the  unTcnown  quantities  on  one  side,  and  the  known 
quantities  on  the  other ;  divide  hy  the  co-efficient  of  the  un- 
known quantity ;  and  extract  the  square  root  of  each  side 
of  the  resulting  equation. 

Solve  the  equation  ?>oi?  —  15  =  0. 

oxr  —  \.b  =^\j  It  -^vill  be  observed  that  the  square  root  of  5 

3:l^=  15  cannot  be  found  exactly,  but  an  approximate 

^  =  5  value  of  it  to  any  assigned  degree  of  accuracy 

•  X  =  -I-a/S  ^^^  ^®  found. 


QUADRATIC   EQUATIONS.  197 

223.  A  root  which  is  indicated,  but  whidi  can  be  found 
only  approximately,  is  called  a  Surd. 

Solve  the  equation  3:r^  +  15  =  0. 

^x  +15  =  0  It  will  be  observed  that  the  square  root 

dx^  =^  —  15  of  —  5  cannot  be  found  even  approximately ; 

x^  z=  —  5  for  the  square  of  any  number,  positive  or 

.  ^  __  _j_-v/Zr5  negative,  is  positive. 

224.  A  root  which   is  indicated,  but  which  cannot  be 
found  exactly  or  approximately,  is  imaginary.     §  206. 

c.  1  Exercise  LXXXIII. 

boive : 

1.  x'-S^AQ.  6.    5x'-d  =  2x'  +  24:. 

2.  2(:i;2-l)-3(5^+l)+14=0.  7.    {x  +  2y==4:X+b.  ^ 

3    ^'-5   ,   2:r^+l^l  g     x"      a:^--10^^^      50  +  r^ 

3  6  2'  *    5  15  25    ' 


1+x      1-x  x'+S  ^+9 

K       3  17  nn     Q      ,   '^       G5^ 

.'•    4^-6^=3-  '''   '"  +  .^-7-- 

^^    4a;2+5  2:r2-5_  7:^2-25 


12. 


10  15              20 

10a;^  +  17  12:r^  +  2^5^^-4 

18  11.-^2-8           9      ' 

14:^-2+16  2:^2+8        2-j2 


13, 

21  8:1^-11        3 


14.    ar^-{-bx-{-a  =  bx  (1  —bx). 
15.    mx^  -\-7i~  q.  16.   x^  ~  ax -\- b  =  ax  (x  —  1). 


198  ALGEBRA. 


Affected  Quadratic  Equations. 

225.  Since  (ax  i  by  =  arj^  i  2  abx  +  P,  it  is  evident  that 
the  expression  a^x^  4: 2  abx  lacks  only  the  third  term,  b^,  of 
being  a  complete  square. 

It  will  be  seen  that  this  third  term  is  the  square  of  the 
quotient  obtained  from  dividing  the  second  term  by  twice  the 
square  root  of  the  first  term. 

226.  Every  aifected  quadratic  may  be  made  to  assume 
the  form  of  a^x^  i  2  abx  =  c. 

The  first  step  in  the  solution  of  such  an  equation  is  to 
complete  the  square ;  that  is,  to  add  to  each  side  the  square 
of  the  quotient  obtained  from  dividing  the  second  term  by 
twice  the  square  root  of  the  first  term. 

■     The  second  step  is  to  extract  the  square  root  of  each  side 
of  the  resulting  equation. 

The  third  and  last  step  is  to  reduce  the  resulting  simple 
equation. 

(1)    Solve  the  equation  16a:^  +  5:r  —  3  =  la^  —  x -,'-  45. 

Simplify,  9ic2  +  6a;=48. 

Complete  the  square,  9x^  +  6a;  +  l=49. 
Extract  the  root,  3x  +  l  =  ±7. 

Reduce,  3a;  =  —  l  +  7or  —  1  —  7, 

3a;=6or-8, 
.-.    rc=2or-2|. 

Verify  by  substituting  2  for  x  in  the  equation 

16ar^  +  5a;  -  3  =  7a;2  -  a;  +  45, 
16  (2f  +  5  (2)  -  3  =  7  (2f  -  (2)  +  45, 
64  +  10-3  =  28-2  +  45, 
71  =  71. 


QUADRATIC    EQUATIONS.  199 

Verify  by  substituting  —  2f  for  x  in  the  equation 

IQ:^  +  5a;  —  3  =  7a^  —  a;  +  45, 
16(-|f  +  5(-|)-3  =  7(-ff -(-■«) +  45; 

1P_24  _  4^0  _  3  _  i4  8   +  I  +  45^ 

1024  -  120  -  27  =  448  +  24  +  405, 
877  =  877. 


(2)    Solve  the  equation  3a:^  —  4:r  =  32. 

Since  the  exact  root  of  3,  the  coefficient  of  a?,  cannot  be  found,  it 
is  necessary  to  multiply  or  divide  each  term  of  the  equation  by  3  to 
make  the  coefficient  of  a:^  a  square  number. 

Multiply  by  3,  9  a;*  -  12  a;  =  96. 

Complete  the  square,    9a;*--12a;  +  4  =  100. 
Extract  the  root,  3  r  —  2  =  ±  10. 

Reduce,  3a;  =  2  +  lOor  2- 10; 

3a;=12or-8. 
.-.  a;  =  4  or  -  2|. 


100 


Or,  divide  by  3, 

^     4a;     32 
3        3' 

Complete  the  square, 

x^- 

4a;     4     32     4     1 
3       9       3      9" 

Extract  the  root, 

3         3 
.    ^_2±10 
3     • 
=  4or-2f. 

Verify  by  substituting  4  for  x  in  the  original  equation, 
48  -  16  =  32, 
32  =  32. 

Verify  by  substituting  —  2f  for  x  in  the  original  equation, 
21i-(-10|)  =  32, 
32  =  32. 


200  ALGEBRA. 


(3)    Solve  tlie  equation  —  3a;^  +  5:r  =  —  2. 

Since  the  even  root  of  a  negative  number  is  impossible,  it  is  necessary 
to  change  the  sign  of  each  term.     The  resulting  equation  is, 


3^2  _ 

■bx  =  2. 

Multiply  by  3, 

9x2-15a;=6. 

Complete  the  square,  9  x^  - 

---f=f- 

Extract  the  root, 

--1=^1- 

Reduce, 

-=¥• 

3a;  =  6  or-l. 

.-.  a!  =  2or-^. 

Or,  divide  by  3, 

,     bx     2 
3       3' 

Complete  the  square,      3^ 

,     5a;     25_49 
3       36      36' 

Extract  the  root, 

6         6 

:.x  =  ^^\ 
6 

=  2or-i 

If  the  equation  3  x^  —  5  a;  =  2  be  multiplied  by /our  times  the  coeffi.- 
eient  of  x^,  fractions  will  be  avoided : 

36a;2-60a;=24. 
Complete  the  square,  3G  a;^  _  go^,  _^  25  =  49. 
Extract  the  root,  6a;  —  5  =  +  7, 

6x=5±7, 
6a;  =12  or -2. 
.-.  x^  2  or  —^. 

It  will  be  observed  that  the  number  added  to  complete  the  square 
by  this  last  method  is  the  square  of  the  coefficient  of  x  in  the  original 
equation  3  a^  —  5  x  =  2. 


QUADRATIC    EQUATIONS.  201 

3  1 

(4)  Solve  the  equation =  2. 

O—X        ZX—D 

Simplify  (as  in  simple  equations), 

4  a;' -23  re  =  -30. 
Multiply  by  four  times  the  coeflficient  of  3?,  and  add  to  each 
side  the  square  of  the  coefficient  of  x, 

64x2  -  ()  +  (23)2  _  529  _  430  =  49. 
Extract  the  root,  8  a;  —  23  =  ±  7. 

Eeduce,  8a;  =  23  ±7; 

8a;  =  30  or  16. 
.-.  a;  =  3|  or  2. 
If  a  trinomial  be  a  perfect  square,  its  root  is  found  by  taking  the 
roots  oiiha  first  and  third  terms  and  connecting  them  by  the  sign  of 
the  middle  term.  It  is  not  necessary,  therefore,  in  completing  the 
square,  to  write  the  middle  term,  but  its  place  may  be  indicated  as  in 
this  example. 

(5)  Solve  the  equation  72^  —  30  a:  =  —  7. 

Since  72  =  2^  X  3',  if  the  equation  be  multiplied  by  2,  the  coeffi- 
cient of  x^  in  the  resulting  equation,  144  a;*  _  60  a;  =  —  14,  will  be  a 
square  number,  and  the  term  required  to  complete  the  square  will  be 
(ff)^"^  (I)' =="¥"■•  Hence,  if  the  original  equation  be  multiplied  by 
4x2,  the  coefficient  of  3?  in  the  result  will  be  a  square  number,  and 
fractions  will  be  avoided  in  the  work. 

Multiply  the  given  equation  by  8, 

576^5 -240  a;  =  -56. 
Complete  the  square,  576  a^  —  (  )  +  25  =  —  31. 
Extract  the  root,  24  a;  —  5  =  ±V— 31. 

Eeduce,  24  a;  =  5  ±  V-31. 

.-.  a;  =  3j\(5±V-31). 
Note.    In  solving  the  following  equations,  care  must  be  taken  to 
select  the  method  best  adapted  to  the  example  under  consideration. 

g  1     .  Exercise  LXXXIY. 

1.  x^-\'4.x-=\2.         4.    :^-7a;  =  8.  7.    a^-x  =  Q>. 

2.  x^~(jx=lQ.         5.    2>x'--4:x=^l.        8.    5:6-2-3a;=2. 

3.  a;2-12a:^^a^*i    a    12^+:r-l=:0.    9.    2a^-21x=U. 


ALQSBIU.. 


4      8«      6  «-l     «-2     «-4 

16.  5«(ay-8)-2(««-0)-»(»  +  8)(af-f4). 

17        8g         5.  83?*  28 

2(»  +  l)     8     ««-l     4(«-iy 

18.  ^-2)(a?-4)-.2(^-l)(«-8)-0. 

19.  i(*-^4)-|(*-2)-.i(2^  +  8). 

ao.  §(8««-«-.6)-i(««-l)-2(«-2)» 

tti    2ag  I    8ag~60  ^12a?-f  70 
15"^8(l04-«)         190     • 

«-1^0      «^^  2»-M     »-.2 

■^  *-l       2x      8'  '"^  7-«^    «        ^ 

M.     2«4-8  7-g    y^7-8a? 

2(2»-l)     2(«  +  l)     4-.8« 

^  12je»-ll««>H0a?>-78^..      ,1 

30^  _e IL.^JL L., 

«  — 1     «  +  5     «+l     «-5' 


QUADRATIO  EQUATIONS.  203 

227.    Literal  quadratic  equations  aro  solved  as  follows : 

(1)  Solvo  tho  of|uation  ai*"  -f  hx  =  c. 

Multiply  tliQ  oquation  by  i  a  and  add  (ho  8quiiro  of  /), 

Extract  thn  root,  2  a»  +  6  -  ±  v^o Jr_6]V^ 

Roduoo,  2  a«  ■•  —  6  i  V4  ae  -f  6*. 

~6dkV4oo-f  6* 

(2)  Solvo  tho  oquation  adx  —  aca^  =  ico:  —  hd. 

TranspoRo  bcx  and  chango  tho  signs, 

aca^  -f  box  —  adx  —  6rf. 
Exproes  llio  h'ft  niombor  in  two  tertM^ 

acx*  ^  (6o  —  ad)  a  —  hd. 
Mnltiply  by  4  ac, 

4  d'c^c"  I-  4  ac  (ft<3  —  aci) « •-  4  aftccf. 
Comploto  Iho  nquaro, 

4  aVx"  +  (  )  +  (6fl  -  at/)*  -  b*c*  \  2  aicci  i  a*d*. 
Extract  Iho  root,     2  oca  +  (5o  -  ad )  -  i  {be  -f-  ad ). 
Reduce,  2  occe  -  -  (6c  -  ad)  i  {bo  i-  ad ) 

-2ador  -2  6c. 

.'.  «  —  -  or  — . 
0         a 

(3)  Solvo  Iho  equation  px*  —  ^u*  -j-  ya;"  -j-  qx  =  -^    . 

Expr<i8H  tho  loft  monibor  in  two  tfrms^ 

pa 


P  «  !? 

Mull  inly  by  four  times  tho  coofliciont  of  a', 
4  (p  +  9)«  «•  -  4  (|)' -  g^) »  -  4;)j. 
Completo  tho  square, 

4  (/)   I  q)*'J^~{  )\-{p-  ?)•-;)*  +  2pq  +  (f. 
Extract  tho  root,  2  (/)  b  g)  a  -  (;>  —  g)  -  db  (/>  l-  g). 
Reduce,  2 (p  +  j) »  -  (j5  -  g)  i:  (/)  i  y), 

-2/>  or  --2g. 

.   ^     _?^o-     _^ 

••     ";>  I  y      "p  M/ 

Note.  Tim  left  Imud  mcimbcr  of  (ho  ocpiaiion  \vh(>n  siniplifiod 
/nuHt  bo  oxprosHod  in  two  terrm,  simple  or  compound,  ono  t(>iiu  con- 
iftining  .)",  !!iii1  ill  >  i>  li(>i-  itM'iii  ('oMiiiiniti''  .r. 


204  ALGEBRA. 


o  1     .  Exercise  LXXXV. 

1.  oc^-\~2ax  =  a?.  14.    7? -{- ax=^a-\-x. 

2.  o?^^ax^na^.  15.   x^ -\- ax  =  hx-\- ah. 

3.  a?=^-^~Zmx.  16.   ^  +  «  =  ^+^ 


4  a     X      b 

4.    ^.-bl^-^^Q,  17.    1  ,       1 


:r 


2  2  :r  '  x  +  h      a  •  a  +  b 

18.    «  +  ^_^  =  0. 


(a;  +  a)2      (a;- a/  3       4       3a 

6.  cx  =  a:^  +  hx^ ^.  19.    ^i+^  =  a  +  ^IH?. 

a  +  6  07-3  a;  +  3 

7.  ^  +  ^^^2a^.  20.    ^^-l^^^O^^^-^^^). 

o''       c-^         c  7nn 

8.  (a^  +  1)  a;  =  aa;^  +  a.  21.    (a:t'  —  Z))  (6a;  —  a)  =  c^. 

o        ah  2  c  ax-[-b      mx  +  w 

y.    -j = .  /5i5. = . 

x  —  a      x  —  b      x  —  c  bx-\-a      7ix-\-m 

iO.    ^ .=  1  +  1  +  1.  23.       ^       ■       ^^ 


a  +  6  +  :r      a      b      x  7n -\-  x      m  —  x 

1^^   _l 1     _^  +  J^      24.    (^-1)^+2(3^-1)^,.^^ 

a  —  x      a-\-x      o?  —  a?-         '  4  a— 1 

^2.   ^^2aM«!±i!)  =  2x       25.    («'-&^)(^^+l)  =  2^. 
a^+6^  a2  +  6^ 

13.      (2^-^0^        5,  26.    ^-^^ff^(m-n)^. 

2:r-a+26  (7?i  +  72)2  ^ 

^      a-5__14a^-5a5-105^  .  (2a- 35):;; 
'   "^  "^    a^>2  ISa^^^  "^        2a6       ' 


Y 


QUADRATIC    EQUATIONS.  205 

28.    ahaf  -\ = ' — . 

c  (T  c 

01?        __  m^  —  4  g^  _  X 
'   Sm-2a     4:a-6m     2' 

30.  Gx-^^ — - — ^  =  5(a  — 6)  +  -— — . 

X  ox 

31.  i(x'  +  a''  +  ab)  =  ix(20a  +  4:b). 

32.  x^  —  (b  —  a^  c^=  ax  —  bx-\-  ex. 

33.  :i;^  — 2ma;  =  (n— ^  +  '^)(^— i^  — ''^)- 

34.  x^  —  (m  -\~  n)  X  =  \  (p  -]-  q  -\-  m  -{-  n)  (p  -{•  q  —  m  —  n). 

35.  m?2a;^  —  (m  +  n)  (mn  +  1)  a;  +  (m  +  w)^  =  0. 

25  —  07  —  2(2  ■  4:b  —  7a_x  —  4:a 
bx  ax  —  bx      ab  ~  b^ 

37.    2x'(a'-  P)  -  (3  a'.+  b')(x-l)  =  (3  5^  +  a')  (x  +  1). 

a  —  2b  —  x        6b  — X     .  2a  — a:  — 195 p. 

(2^  —  45^        aa;  +  25:r         25^7  — aa; 

39. 


^  rg  +  13a  +  35      ■^^_a-26 

'  ba-Sb-x  x  +  2b' 

07  +  35 35 ^  +  35  _^ 

'  8a^-12ab      9b^-4:a^      (2a  +  35)(o7-35) 

41.  net? -{-px  —pc?  —  mx -\-7n  —  n^=^(). 

42.  {a-\-b  +  c)x^-{2a-\-b  +  c)x  +  a=^0. 

43.  (ax —  5)  {c  —  d)^={a  —  5)  (ex  —  d)  x. 
^  2x-\-\      I  fl      2\_3o;+l 

45. 


5  o;  \5 

1  ,  1  _       «  25o7  +  5 


23i?  +  x-\      2x^-^x^\      2bx-b       ax^-a' 


206  ALGEBRA. 


228.  An  affected  quadratic  may  be  reduced  to  the  form 
x^  -{-px-\-  q=^0,  in  which  p  and  q  represent  any  numbers, 
positive  or  negative,  integral  or  fractional. 

Ex.  Solve:  s^ -\-px-\- q  =  Q. 

2x  -{-p  =  zb  -Vp^  —  4  5', 

JBy  this  formula,  the  values  of  x  in  an  equation  of  the 
form  x?-\-px-^q^^,  may  be  written  at  once.  Thus,  take 
the  equation 

Divide  by  3,      ar'  -  f  x  +  |  =  0. 

Here,  P  =  —  f.  ^-nd  g'  =  f . 

=  1  or  |. 

229«  A  quadratic  which  has  been  reduced  to  its  simplest 
form,  and  has  all  its  terms  written  on  one  side,  may  often 
have  that  side  resolved  hy  inspection  into  factors. 

In  this  case,  the  roots  are  seen  at  once  without  com- 
pleting the  square. 

(1)    Solve  a;2_^  7^  _  60  =  0. 

Since  ar^  +  7a;  -  60  =  (a;  +  12)  {x  -  5), 

the  equation  x^  +  7a;  —  60  =  0 

may  be  written    {x  +  12)  (a;  —  5)  =  0. 

It  will  be  observed  that  if  dilier  of  the  factors  a;  +  12  or  a;  —  5  is  0, 
'Cq.^ 'product  of  the  two  factors  is  0,  and  the  equation  is  satisfied. 

Hence,  a;  +  12  =  0  and  a;  —  5  =  0. 

:.  X  —  —  12.  and  a;  =  5. 


QUADRATIC   EQUATIONS.  207 

C2)    Solver?;2_|_7^^0. 

The  equation  a:*  +  7a:  =  0 

becomes  x(a;  +  7)  =  0, 

and  is  satisfied  if  a;  =  0,  or  if  a;  +  7  =  0. 

.•.  the  roots  are  0  and  —  7. 

It  will  be  observed  that  this  method  is  easily  applied  to  an  equation 
all  the  terms  of  which  contain  x. 


(3)    Solve  2a;3-a;^- 6a:  =  0. 

The  equation  2a^  — a^  —  6a;  =  0 

becomes  a;  (2  x'  —  a;  —  6)  =  0, 

and  is  satisfied  if  ;r  =  0,  or  if  2  a:.-^  —  x  ~  6  =  0. 

By  solving  2x^  —  a;— 6  =  0  the  two  roots  2  and  —  f  are  found. 
.*.  the  equation  has  three  roots,  0,  2,  —  f . 


(4)    Solve  a:3  +  a,-2  — 4^-4  =  0. 

The  equation  a^  +  a;'  —  4x  —  4  =  0 

becomes  a:^  (a;  +  1)  —  4  (x  +  1)  =  0, 

(a,'*-4)(a;  +  l)  =  0. 
.'.  the  roots  of  the  equation  are  —  1,  2,  —  2. 


(5)    ^o[\ea^~2x^-llx+l 


Since  ^-ZL^^l=Lil^±_i^^^_^_12, 

x—\ 

the  equation  a:^-2ar^-lla;  +  12  =  0 

may  be  written  {x  -  1)  (a,^  —  a;  —  12)  =  0. 

The  three  roots  are  found  to  be  1,  —  3,  4. 

An  equation  which  cannot  be  resolved  into  factors  by  inspection 
may  sometimes  be  solved  by  guessing  at  a  root  and  reducing  by  divi- 
sion. In  this  case,  if  a  denote  the  root,  the  given  equation  (all  the 
terms  of  the  equation  being  written  on  one  side),  may  be  divided  by 


208  ALGEBRA. 

Exercise  LXXXVT. 
Find  the  roots  of: 

1.  (x+l)(x-2)(x^+x-2)=0.  7.  x^--x^--x  +  l=0. 

2.  (x'-3x-i-2')(x'-x-12)=0.         8.  8:^3-1  =  0. 

3.  (x+l)(x-2)(x+S)  =  -(j.  9.  8:^^+1=0. 

4.  2a^  +  4:x^~70x  =  0.  10.  :r^-l=0. 

5.  (x'~x-6)(x'-x-20)  =  0.  11.  x(x-a)(x^-b')^0. 

6.  x(x+l)(x+2)=(a-^2')(a+l)a.  12.  n(3^+l)-\-x+l  =  0. 

230.    If  r  and  /  represent  two  values  of  x,  then 

x  —  r  =  0, 
and     .  x  —  r'  =  0, 

.'.  {x  —  r)  (:r  —  /)  =  0. 

This  is  a  quadratic  equation,  as  may  be  seen  by  performing  the 
indicated  multipUcation. 

Now  r  and  r'  are  roots  of  this  equation ;  for,  if  either  r  or  r'  be 
written  for  ic,  one  of  the  factors,  x  —  r,  x~r',  is  equal  to  0,  and  the 
equation  is  satisfied.  Also  r  and  r'  are  the  only  roots,  for  no  value 
of  X,  except  r  and  r',  can  make  either  of  these  factors  equal  to  0. 

Since  r  and  r'  may  represent  the  values  of  x  in  any  quadratic 
equation,  it  follows  that  every  quadratic  equation  has  two  roots,  and 
only  two. 

Again,  if  r,  r',  r'',  represent  three  values  of  x, 
then,        .  {x-r)lx-  r')  (x  -  r")  =  0. 

This  is  a  cubic  equation,  as  may  be  seen  by  performing  the  indi- 
cated multiplication.  Hence,  it  may  be  inferred  that  a  cubic  equa- 
tion has  three  roots,  and  only  three ;  and  so,  for  any  equation,  that 
the  number  of  roots  is  equal  to  the  degree  of  the  equation. 

It  may  also  be  inferred  that  if  r  be  a  root  of  an  equation,  x  —  r 
will  be  a  factor  of  the  equation  when  the  equation  is  written  with  all 
its  terms  on  one  side. 


QUADRATIC   EQUATIONS.  209 

If  r  and  r'  represent  the  roots  of  the  general  quadratic  equa- 
tion, x^  +  px  +  2-  =  0. 

This  equation  may  be  written   {x  —  r){x  —  r')  =  0, 
or,  a^—  {r  +  r')  x  +  rr'  =  0. 

A 'form  which  shows  that 
the  sum  of  the  roots  »=»  —p, 

and  the  product  of  the  roots  =  q. 

231.  It  will  be  seen  from  §230  that  an  equation  may  be 
formed  if  its  roots  be  known. 

If  the  roots  of  an  equation  be  —  1  and  \, 
the  equation  will  be  (x  -\-V){x  —  \)  =  0, 

or.  a^4-§^-l  =  0, 

4       4 

or,  by  multiplying  by  4,  4  ar*  +  3  a;  —  1  =  0. 

If  the  roots  of  an  equation  be  0,  1,  5, 
the  equation  will  be  (a;  —  0)  (x  —  1)  (a;  —  5)  =  0 ; 

that  is,  x{x  —  Hi^  —  5)  =  0, 

or,  a^  — .6a^  +  5a;  =  0. 

If  X  occur  in  every  term  the  equation  will  be  satisfied  by  putting 
ar  =  0,  and  may  be  reduced  to  an  equation  of  the  next  lower  degree 
by  dividing  every  term  by  x. 

232.  By  considering  the  roots  of  o^  -\-px  -\-  q  =  0, 
namely,  r  =  — ^  +  -  -y/p^  -Aq, 

and  /=—-£_- yp34^, 

it  will  be  seen  that  the  character  of  the  roots  of  an  equation 
may  be  determined  without  solving  it : 

I.  As  the  two  roots  have  the  same  expression,  V^^  ~  4  5", 
both  roots  will  be  ^al,  or  both  will  be  imaginary/. 

If  both  be  real,  both  will  be  rational  or  both  surds,  accord- 
ing as  p^  —  Aq  is  or  is  not  a  perfect  square. 


210  ALGEBRA. 


II.  When  jp^  is  greater  than  4  q,  the  two  roots  will  be  real, 
for  then  the  expression  p^  —  ^^q  is  positive,  and  therefore 
-\/p^  —  4:q  can  be  found  exactly  or  approximately. 

Since  also  its  value  in  one  root  is  to  be  added  to  —  ^, 

and  in  the  other  to  be  subtracted  from  — -,   the  two  roots 
will  be  different  in  value. 

III.  When  p^  is  equal  to  Aq,  the  roots  will  be  equal  in 
value. 

IV.  When  p^  is  less  than  4  q,  the  roots  will  be  imaginary, 
for  then  the  expressions^— 4  5*  will  be  negative,  and  therefore 
Vp^  —  4  ^  represents  the  even  root  of  a  negative  number,  and 
is  imaginary. 

V.  If  2'  (='^  X  /)  be  positive,  the  roots,  if  real,  will  have 
the  same  sign,  but  opposite  to  that  oi  p  (since  r  +  /  =  —p). 

But  if  q  be  negative,  the  roots  will  have  opposite  signs. 

233.  Determine  by  inspection  the  character  of  the  roots  of: 

(1)  ar'-bx  +  Q^O. 

In  this  equation  p  is  —d,  and  §•  is  6. 

.-.  \/p''-4^=V25-24  =  1. 
.*.  the  roots  will  be  rational,  and  both  positive. 

(2)  x^  +  3x-\-l=0. 

In  this  equation,  p  is  3,  and  q  is  1. 


.-.  Vp^-iq  =V9-4=\/5. 
the  roots  will  be  Surds,  and  both  negative. 


r3)    x^-{-Sx^4:  =  0. 

In  this  equation  p  is  3,  and  q  is  4. 


.-.  Vp^~4:q==V9-16 
y  .'.  the  roots  will  be  hnpossiblc 


quadratic  equations.  211 

Exercise  LXXXVII. 
Form  the  equations  whose  roots  are : 

1.  2,  1.  5.   -5,-f  9.   0,-i  1,-1. 

2.  7,-3.  6.   -f,f  10.   a-2^»,3a  +  25. 

3.  i   |.  7.    3, -3,  f, -f.      11.   2a-h,h-?>a. 

4.  I, -f.  a   0,1,2,3.  12.   a(a+l),  1-a. 
Determine  by  inspection  the  character  of  the  roots  of : 

13.  ^-7a;  +  12  =  0.  17.  x^-^4:X-{-l  =  0. 

14.  x--1x  —  2>0  =  0.  18.  :i'2-2a;  +  9  =  0. 

15.  oc^-{-^x—b  =  0.  19.  3:r2-4a:-4  =  0. 

16.  bx^  +  ^=^0.  20.  a;2  +  4a7  +  4  =  0. 

234.    It  is  often  useful  to  determine  the  maximum  or  mini- 
mum value  of  a  given  quadratic  expression. 

(1)    Find  the  maximum  or  minimum  value  of  1  +  a;  —  :r^. 
Let  1  -\-  X  —  2^  =  m] 

then,  a^  —  re  =  1  —  m, 

and  4j::*  — 0  +  1  =5  — 4m, 

2a;— 1  =  ±V5  — 4m. 


j;.  a;=  J±^  v5  — 4m. 
Now,  for  all  possible  values  of  a;,  5  — 4m  cannot  be  negative;  that 
is,  m  cannot  be  greater  than  f ;  and  for  this  value  x  is  ^.     Therefore, 
f  is  the  maximum  value  of  the  given  expression. 

(2)    Find  the  maximum  or  minimum  value  of  :c^+  3  a:  +  4. 
Let  a;'  +  3a;  +  4  =  m; 

then,  a;*  +  3  a;  =  m  —  4, 

and,  43;*  +  (  ) +  9  =  4m  — 7, 

2a;  +  3  =  ±V4m-7. 

a;  =  —  I  ±  J  V4  m  —  7. 
For  all  possible  values  of  a;,  4m  — 7  cannot  be  negative;  that  is, 
m  cannot  be  less  than  | ;  and  for  this  value  a;  =  —  f .     Therefore,  |  is 
the  minimum  value  of  the  given  expression. 


212  .  ALGEBRA. 


Exercise  LXXXVni. 

Find  the  maximum  or  minimum  value  (and  determine 
which)  of: 
1.    4  +  6a;  — :r2.         4.    (a-x){x-h).        7.   x^—2x-{-9. 


2.    (^  +  ^y  5.    _^ 


5.    — ^^^.  8. 


X  l-f-x^  (x-\-a)(x—b) 

3.    ^!±1.  6.   ^  +  8a;+2a         9.   — ^. 

^  a  +  ar 

10.  Divide  a  line  20  in.  long  into  two  parts  so  that  the  sum 

of  the  squares  on  these  two  parts  may  be  the  least 
possible. 

11.  Divide  a  line  20  in.  long  into  two  parts  so  that  the  rect- 

angle contained  by  the  parts  may  be  the  greatest 
possible. 

12.  Find  the  fraction  which  has  the  greatest  excess  over  its 

square. 

235.    Two  other  cases  of  the  solution  of  equations  hi/  com- 
pleting the  square  should  be  noticed. 

I.  When  any  two  powers  of  x  are  involved,  one  of  which 
is  the  square  of  the  other.  * 

II.  When  the  addition  of  a  number  to  an  equation  of  the 
fourth  degree  will  make  both  sides  complete  squares. 

(1)    Solve8:?;«+63.T3=8. 

In  this  equation  the  exponent  G  is  the  double  of  3,  hence  x^  is 
the  square  of  s?. 

Sa-s  +  GSa^z^S, 
256a^  +  () +(63)2  =  4225, 
lGa^  +  G3  =  ±65, 

lG:t3  =  2,  or-128, 
2,3  =  i  or  -  8. 
By  taking  cube  root,  a;  =  J,  or  —  2. 


QUADRATIC   EQUATIONS. 


213 


The  other  roots  of  the  equation  are  found  by  finding  the  remain- 
ing  roots  of  the  equations,  rc^  =  |,  and  a^  =  —S. 


Since,  a:^^^^  .-.  8a;3-l  =  0 

Now,  by  I  230, 

8x3  -  1  =  (2a;  -  i)  ^4^^  +  2a;  +  1) 

.•.(2a;-l)(4a^  +  2a;  +  l)  =  0 

and  is  satisfied  if4x'  +  2a;  +  l  =  0 

as  well  as  if  2  a;  — 1  =  0 

The  solution  of4x'  +  2a;  +  l  =  0 
gives      a;  =  :^  (—  1  ±  V^^). 


Since,  a;^  =  —  8,       .-.  a;'+8  =  0 

Now,  by  g  230, 

a;3  +  8  =  (a;  +  2)  (a;«  -  2a;  +  4) 

.•.(a;  +  2)(ar'.-2a;  +  4)  =  0 

and  is  satisfied  if  a-^  —  2a;  —  4  =  0 

as  well  as  if  x  +  2  =  0 

The  solution  of  x^  -  2  a;  +  4  =  0 
gives  x=l  ±V^^. 


.'.  the  roots  are  |,  —  2,  1  ±  V—  3,  |^  (—  1  ±  V— 3). 

(2)    Solve  :r*-10a;3  + 35:^^ -50a; +  24  =  0. 
Take  the  square  root  of  the  left  side. 


2x'-6x 


-10^  +  35:^2 

-10x^-h25x^ 


2a;2-i0a;4-5 


10^- 50a; +  24 
10^-50^  +  25 


-    1 


It  is  now  seen  that  if  1  were  added  the  square  would  be  complete, 
and  the  equation  would  be 

X*  _  10 :?.-3  +  35  a^  -  50x  +  25  =  1. 

Extract  the  square  root  and  the  result  is, 

a;^  —  5a;  +  5  =  ±l, 
That  is,  x2  -  5  a;  =  -  4,  or  -  6, 

4a;2_()  +  25  =  9,  orl, 

2  a;  —  5  =  ±  3,  or  ±  1. 
2a;  =  8,  2,  6,  or  4, 
.-.  a;  -  4,  1,  3,  or  2. 


214  ALGEBRA. 


Exercise  LXXXIX. 
Find  the  possible  roots  of: 

1.  3^+7x^  =  8.  8.  (a^-9y  =  S  +  ll(x''-2). 

2.  x*  —  5a^  +  4:  =  0.  9.  x^-\-14:a^-{-24:  =  0. 

3.  S7a^-9==4:x\  10.  19a;*  +  216a;^  =  a7. 

4.  165;8=17:r*-l.  11.  5;H  22a;* +  21-0. 

5.  32ari°-33:r^+l=:0.  12.  a;^"*  +  3  a;"*  -  4  =  0. 

6.  (:r2-2)2=}(a;2_^12).  13.  4:^^-20 ^+23:^2+5 a; =6. 

7.  ^4n_5^_25^Q  j^^     1   _|.1_20  =  0. 

3         12  :c2"      a;** 

15.  x*-4:x^~103^  +  28x-15  =  0. 

16.  a;*-2x'3-13a;2+14a;  =  -24. 

17.  108:r*  =  20a:(9:r2-l)-51a;2_f_7_ 

18.  (xr'-l)(a^-2)  +  (x^-S){a^-4:)  =  x*  +  6. 

Problems  Involving  Quadratics. 

236.  Problems  which  involve  quadratic  equations  have 
aiDparently  two  solutions,  as  a  quadratic  has  two  roots. 
Sometimes  both  will  be  solutions ;  but  generally  one  only 
will  be  a  solution,  and  the  other  be  inconsistent  with  the 
conditions  of  the  problem.  No  difficulty  will  be  found 
in  selecting  the  result  which  belongs  to  the  problem,  and 
sometimes  a  change  may  be  made  in  the  statement  of  a 
problem  so  as  to  form  a  new  problem  corresponding  to  the 
solution  which  was  inapplicable  to  the  original  problem. 


QUADRATIC   EQUATIONS.  215 

(1)  The  sum  of  the  squares  of  two  consecutive  numbers  is 
481.     Find  the  numbers. 


Let 

X  =  one  number, 

and 

X  +  1  =  the  other. 

Then 

a^  +  {x  +  lf  =  481, 

or 

2j^  +  2a;  +  1  =  481. 

The  solution  of  which  gives,     x  =  15,  or  —  16. 

The  positive  root  15  gives  for  the  numbers,  15  and  16. 

The  negative  root  —  16  is  inapplicable  to  the  problem,  as  consecu- 
tive numbers  are  understood  to  be  integers  which  follow  one  another 
in  the  common  scale,  1,  2,  3,  4 • 

(2)  What  is  the  price  of  eggs  per  dozen  when  2  more  in  a 
shilling's  worth  lowers  the  price  1  penny  per  dozen  ? 
Let  X  =  number  of  eggs  for  a  shilling. 

Then,  -  =  cost  of  1  egg  in  shillings, 

X 

12 

and  —  =  cost  of  1  dozen  in  shillings. 

But,  if  X  +  2  =  number  of  egga  for  a  shilling, 

12 

cost  of  1  dozen  in  shillings. 


x  +  2 
12        12         1 


(1  penny  being  j^  of  a  shilling). 


X      x  +  2     12 
The  solution  of  which  gives  x  =  16,  or  — 18. 

And,  if  16  eggs  cost  a  shilling,  1  dozen  will  cost  ||  of  a  shilling, 
or  9  pence. 

Therefore,  the  price  of  the  eggs  is  9  pence  per  dozen. 

If  the  problem  be  changed  so  as  to  read :  What  is  the 
price  of  eggs  per  dozen  when  two  less  in  a  shilling's  worth 
raises  the  price  1  penny  per  dozen  ?  the  algebraic  statement 
will  be  J^_12^1_ 

x-2      X       12' 

The  solution  of  which  gives  x  =  18,  or  —  16. 

Hence,  the  number  18,  which  had  a  negative  sign  and  was  inappli- 
cable in  the  original  problem,  is  here  the  true  result. 


216  ALGEBRA. 


Exercise  XC. 

1.  The  sum  of  the  squares  of  three  consecutive  numbers  is 

365.     Find  the  numbers. 

2.  Three  times  the  product  of  two  consecutive  numbers 

exceeds  four  times  their  sum  by  8.    Find  the  numbers. 

3.  The  product  of  three  consecutive  numbers  is  equal  to 

three  times  the  middle  number.     Find  the  numbers. 

4.  A  boy  bought  a  number  of  apples  for  16  cents.     Had 

he  bought  4  more  for  the  same  money  he  would  have 
paid  i  of  a  cent  less  for  each  apple.  How  many  did 
he  buy  ? 

5.  For  building  108  rods  of  stone- wall,  6  days  less  would 

have  been  required  if  3  rods  more  a  day  had  been 
built.     How  many  rods  a  day  were  built  ? 

6.  A  merchant  bought  some  pieces  of  silk  for  ^  900.     Had 

he  bought  3  pieces  more  for  the  same  money  he  would 
have  paid  $15  less  for  each  piece.  How  many  did 
he  buy?  /;i^ 

7.  A  merchant  bought  some  pieces  of  cloth  for  $168.75. 

He  sold  the  cloth  for  $  12  a  piece  and  gained  as  much 
as  1  piece  cost  him.  How  much  did  he  pay  for  each 
piece  ? 

8.  Find  the  price  of  eggs  per  score  when  10  more  in  622 

cents'  worth  lowers  the  price  31  i  cents  per  hundred. 

9.  The  area  of  a  square  may  be  doubled  by  increasing  its 

length  by  6  inches  and  its  breadth  by  4  inches.  De- 
termine its  side. 

10.  The  length  of  a  rectangular  field  exceeds  the  breadth 
by  1  yard,  and  the  area  is  3  acres.  Determine  its 
dimensions. 


QUADRATIC   EQUATIONS.  217 

11.  There  are  three  lines  of  which  two  are  each  ^  of  the 

third,  and  the  sum  of  the  squares  described  on  them 
is  equal  to  a  square  yard.  Determine  the  lengths  of 
the  lines  in  inches.        f  ^  ^^  (^  ^  ^  p 

12.  A  grass  plot  9  yards  long  and  6  yards  broad  has  a  path 

round  it.  The  area  of  the  path  is  equal  to  that  of 
the  plot.     Determine  the  width  of  the  path. 

13.  Find  the  radius  of  a  circle  the  area  of  which  would  be 

doubled  by  increasing  its  radius  by  1  inch. 

14.  Divide  a  line  20  inches  long  into  two  parts  so  that  the 

rectangle  contained  by  the  whole  and  one  part  may 
be  equal  to  the  square  on  the  other  part. 

15.  A  can  do  some  work  in  9  hours  less  time  than  B  can  do 

it,  and  together  they  can  do  it  in  20  hours.  How 
long  will  it  take  each  alone  to  do  it  ? 

16.  A  vessel  which  has  two  pipes  can  be  filled  in  2  hours 

less  time  by  one  than  by  the  other,  and  by  both  to- 
gether in  2  hours  55  minutes.     How  long  will  it  take 
,  each  pipe  alone  to  fill  the  vessel  ? 

17.  A  vessel  which  has  two  pipes  can  be  filled  in  2  hours 

less  time  by  one  than  by  the  other,  and  by  both  to- 
gether in  1  hour  52  minutes  30  seconds.  How  long 
will  it  take  each  pipe  alone  to  fill  the  vessel  ? 

18.  An  iron  bar  weighs  36  pounds.     If  it  had  been  1  foot 

longer  each  foot  would  have  weighed  I  a  pound  less. 
Find  the  length  and  the  weight  per  foot. 

19.  A  number  is  expressed  by  two  digits,  one  of  which  is  the 

square  of  the  other,  and  when  54  is  added  its  digits 
are  interchanged.     Find  the  number. 

20.  Divide  35  into  two  parts  so  that  the  sum  of  the  two 

fractions  formed  by  dividing  each  part  by  the  other 
may  be  2-^-^. 


218  ALGEBRA. 

21.  A  boat's  crew  row  3  J  miles  down  a  river  and  back 
again,  in  1  hour  40  minutes.  If  the  current  of  the 
river  i«  2  miles  per  hour,  determine  their  rate  of  row- 
ing in  still  water.  < 

32.  A  detachment  from  an  army  was  marching  in  regular 
column  with  5  men  more  in  depth  than  in  front.  On 
approaching  the  enemy  the  front  was  increased  by 
845  men,  and  the  whole  w^as  thus  drawn  up  in  5  lines. 
Find  the  number  of  men. 

23.  A  jockey  sold  a  horse  for  $  144,  and  gained  as  much  per 

cent  as  the  horse  cost.     What  did  the  horse  cost  ? 

24.  A  merchant  expended  a  certain  sum.  of  money  in  goods, 

which  he  sold  again  for  $  24,  and  lost  as  much  per 
cent  as  the  goods  cost  him.  How  much  did  he  pay 
for  the  goods  ? 

25.  A  broker  bought  a  number  of  bank  shares  ($100  each), 

when  they  were  at  a  certain  per  cent  discount,  for 
$  7500  ;  and  afterwards  when  they  were  at  the  same 
per  cent  premium,  sold  all  but  60  for  ^5000.  How 
many  shares  did  he  buy,  and  at  what  price  ? 

26.  The  thickness  of  a  rectangular  solid  is  I  of  its  width, 

and  its  length  is  equal  to  the  sum  of  its  width  and 
thickness;  also,  the  number  of  cubic  yards  in  its  vol- 
ume added  to  the  number  of  linear  yards  in  its  edges 
is  I"  of  the  number  of  square  yards  in  its  surface. 
Determine  its  dimensions. 

27.  If  a  carriage-wheel  16  J  feet  round  took  1  second  more 

to  revolve,  the  rate  of  the  carriage  per  hour  ^vould  be 
H  miles  less.    At  what  rate  is  the  'iarriage  travelling? 


CHAPTER    XV. 

Simultaneous  Quadratic  Equations. 

^37.  Quadratic  equations  involving  two  unknown  quan- 
tities require  different  methods  for  their  solution,  according 
to  the/orm  of  the  equations. 

238.  Case  I.  "When  from  one  of  the  equations  the  value 
of  one  of  the  unknown  quantities  can  be  found  in  terms  of 
the  other,  and  this  value  siibstituted  in  the  other  equation. 

Ex.    Solve:  3^-2.y  =  5|  (1) 

x-y  =  2  J  (2) 

Transpose  x  in  (2),  y  =  x  —  2. 

Substitute  in  (1),  3  a*  -  2a;  (x  -  2)  =  5. 

The  solution  of  which  gives  a;  =  1  or  —  5. 


y 


1  or  -  7. 


Special  methods  often  give  more  elegant  solutions  of  examples  than 
the  general  method  by  substitution. 

I.  When  equations  have  the  form,  x  ±  y  =  a,  and  xy  =  b ;  x^  ±  y^ 
=  a,  and  xy  ==  b ;  or,  x  ±  y  =  a,  and  x^  +  y'  =  6. 

W    Solve:  "  +  ^o^.''l  ^o! 

V^  a:y  =  300     J  (2) 

Square  (1),  a^  +  2  xy  +  /  =  1600.  ^3) 

Multiply  (2)  by  4,  4xy         =1200.  (4) 

Subtract  (4)  from  (3),  x^-2xy+y^  =  400. 

Extract  root  of  each  side.  x  —  y  =  ±  20.  (6) 

Add  (1)  and  (6),  2  x  =  60  or  20, 

.•.x  =  30or  10. 
Subtract  (6)  from  (1),  2y  =  20  or  60, 

,'.  y  =  10  or  30. 


220  ALGEBRA. 


Square  (1),  x^ -2xy +y^=^lQ.  (3) 

Subtract  (2)  from  (3),  -  2  rcy  =  -  24.  (4) 

Subtract  (4)  from  (2),  ar^  +  2  xy  +  /  =  64. 

Extract  the  root,  x  +  y  =  ±  8.  (5) 

By  combining  (5)  and  (1),  a;  =  6  or  —  2. 

y  =  2  or  -  6. 


(3)    Solve: 


(1) 
(2) 


Square  (1),  i+A+I      il.  (3) 

^         ^  ^  0?     xy     y«     400  ^  ^ 

Subtract  (2)  from  (3),  A  =  i?..  (4) 

xy     400 

Subtract  (4)  from  (2).       i  _  A  +  i  =.  J_. 
^  ^  ^  ^'       :^     xy     y''     400 

Extract  tbe  root,  =  +  —  (5) 

a;     y      "20  ^  ' 

By  combining  (1)  and  (5),  x  =  4  or  5. 

y  =  5  or  4. 


II.    When  one  equation  may  he  simplified  by  dividing  it  by  the  other. 

(4)    Solve:  7        7       f  ^9^ 

Divide  (1)  by  (2),  x^~xy  +  y^=  13.  (3) 

Square  (2),  a,-''  +  2  a:y  +  y"  =  49.  (4) 

Subtract  (3)  from  (4),  3  xy  =  36. 

Divide  by  -  3,  -xy  =  - 12.  (5) 

Add  (5)  and  (3),  a.-^  -  2  zy  +  y'^  -  1. 

Extract  the  root,  x  —  y  =  ±  1.  (6) 

By  combining  (6)  and  (2),  a;  =  4  or  3. 

V  =  3  or  4. 


SIMULTANEOUS    QUADRATIC   EQUATIONS. 


221 


Solve : 

1.  :r  +  y  =  13| 
a;y  =  36         J 

2.  :r  +  y  =  291 
xy  =  100     J 

3.  :r-y=19| 
a:y  =  66       J 

4.  a:  —  y  =  45  ) 

.ry  =  250     J 


Exercise  XCI. 


11.  a;  +  y  =  49        | 

12.  r»  +  3/*  =  341'l 
a:  +  y  =  ll      J 

13.  :r^  +  y^  =  10081 
x-\-y=l2        J 

14.  a;^-?/'  =  98) 
a:-y  =  2      J 


5.   a;  — y  =  10      I 

:r2  +  y2=178j 


15.    a,-»-y»  =  279) 
a;-y  =  3        i 


6.   a;  —  y  =  14      \ 
=  436  J 


^  +  y' 

x-{-y  = 
x^^f 


7.   a;  +  y  =  12      | 
=  104  J 


a  1+1=1 

x      y      ^ 

1    ,  i^_5 
a:2"^y2      16  J 


a  1+1=5 
^     y 

Vi=13 

10.    7r'-8:ry=1591 
5a:  +  2y  =  7        J 


16. 

^-3y  =  n 
^y  +  y2  =  5J 

17. 

4y  =  5:r+l    ) 
2^y  =  33-;r2/ 

18. 

1-1=3      1 

X      y 

■ 

19. 

X      y 

^      y^ 

20. 

x^-2xy^f 
x  +  y  =  2 

=  1 

222  ALGEBRA. 


239.    Case  II.    When  each  of  the  two  equations  is  homo- 
geneous  and  of  the  second  degree. 

Let  y  =  vx,  and  substitute  vx  for  y  in  both  equations. 

From(l),        2v2^2_4^^  ^  3^_  17^ 


From  (2), 

Equate  the  values  of  x', 


■.  x2  = 

17 

2v'- 

-4t; 

+ 

3 

r^x"- 

-0?  = 

16, 

■.^  = 

16 

T- 

17 

16 

2t''-4v  +  3      v'-V 

32i;2-61v  +  48=  17^;''-17, 

15t;2-64v= -65. 


The  solution  gives, 

„      5  ^^  13 

Substitute  the  value  of  v  in 

-%-^' 

then, 

—  f 

.■ 

o            5 

.'.  a;=  ±  3    or  ±-, 
o 

and 

13 
y  =  vx  =  ±  5  or  ±  — . 
3 

„  1  Exercise  XCII. 

Solve : 

1.  a^  +  :r2/  +  22/'=74    1 
2a;2  +  2a:y  +  3/2^73/ 

2.  ar^  +  :cy  +  42/2^e| 
+  87/ =  14      J 


3r^  +  82/2 

y^  —  2:ry  =  — 15 


3.   :^-xy  +  f  =  2l}^ 


4. 

.r2-4?/2- 

-9  = 

=  0/ 

* 

:ry  +  2y2- 

-3- 

5. 

^^-^y- 

35  = 

=  0| 

=  0/ 

^7  +  y^  - 

18  = 

6. 

'x'  +  xy-Y 

■V 

=  44| 
=  16  J 

'^ 

2x^-xy  +  y'' 

'' 

^^ 

SIMULTANEOUS    QUADRATIC   EQUATIONS.  )  223' 

7.   cc^  +  xT/-15  =  0)  9.    2:r2  +  3a;y  +  2/2  =  70 

-7/-2  =  0    i 


ar'  —  x?/^7/=7  )     10.    x^  —  x7j  —  y^  =  5 

Sx"  +  13x7/ +  8y'=  162)  27?  ^3xy -\-f  =  28 


240.   Case  III.  When  the  two  equations  are  symmetrical 

with  respect  to  x  and  y ;  that  is,  when  they  have  x  and  y 
similarly  involved  in  them. 

Thus,  the  expressions  2a^  +  3 ar^y'  +  23/^,  2a;y  —  3x  —  3?/  +  ],  a;'*-^ 
3x^2/  —  Sxy^  +  3/^  are  symmetrical  expressions. 

(1)  Solve:  =?  +  f  =  lSxy^  (1) 

x  +  y  =  \2        f  (2) 

Put  w  +  V  for  re,  and  m  —  v  for  y,  in  (1)  and  (2). 

(1)  becomes        {u  +  vf  +  (u  ~  vf  =  18  (m  +  v)  [u  -  v), 

or  u^  +  3uv'  =  9(u^-v^).  (3) 

(2)  becomes  {u  +  v)  -f-  (u  —  v)  =  12, 

or  2m  =  12, 

.-.  M  =  6. 
Substitute  6  for  u  in  (3). 

(3)  becomes  216  +  ISv"  =  9  (36  -  v% 
whence,  i^  =  4:, 

.-.v  =  ±  2, 

.-.x  =  u  +  v  =  6  ±  2  =  8  or  4, 
and  y  =  M  —  v  =  6T2  =  4or8. 

(2)  Solve:  ^  +  y  =  8        |  (1) 

X*  +  y*=  706  J  (2) 

Put  w  +  V  for  X,  and  m  —  v  for  y,  in  (1)  and  (2). 

(1)  becomes  (u  +  r)  +  (u  —  v)  =  8, 

.'.  M  =  4. 

(2)  becomes  w<  +  6w2y2  +  -y*  =  353.  (3) 
Substitute  4  for  u  in  (3), 

256  +  96^r«  +  i;^  =  353, 
or,  0,4  4.  96^5  _  97  ^^^ 

The  solution  of  (4)  gives  v  =  ±  1  or  ±  V—  97. 

Taking  the  possible  values  of  v,  re  =  5  or  3,  and  y  =  3  or  5. 


224  ALGEBRA. 

o  1  Exercise  XCIII. 

boive  : 

1.  4^y  =  96— :r2/|  4.   A:{x-\-y)  =  ?>xy         1 
a7  +  y  =  6            J  a;  +  y  +  ^^  +  y^  =  26/ 

2.  a:2_|_^^ig_^„^|        5^    4:i'2  +  a;y  +  4?/2  =  58  I 

3.  2{:^  +  y')=--bxy\  6.    :ry  (;r  +  y)  =  30  | 
4  (a;  —  y)  =  0:3/      J  5;^  +  y^  =  35        J 

241.  The  preceding  cases  are  general  methods  for  the  solution  of 
equations  which  belong  to  the  kinds  referred  to ;  often,  however,  in 
the  solution  of  these  and  other  kinds  of  simultaneous  equations  in- 
volving quadratics,  a  little  ingenuity  will  suggest  some  step  by  which 
the  roots  may  easily  be  found. 

CI  1  Exercise  XCIV. 

Solve  : 

1.  x-y=*l  \  8.   x--y  =  l      1 
x'  +  xy  +  y'  =  l^}  a^  +  y'==8U 

2.  a^  +  xy  =  S5)  9.   x^  +  4:xy=S    | 
xy-f  =  6    J  4:ry  +  y2=2iJ 

3.  xy-12  =  0^  10.   x^-xy  +  y^--=4:8'\ 
x-2y  =  b   J  x--y~8  =  0        J 

4.  xy-1  =  0    1  11.   x'  +  Sxyi-y^=l      ) 
x'  +  y^  =  50)  3:r2  +  .^y  +  3y2  =  13/ 

5.  2^-5y  =  9        1  12.   a^-2xy  +  Sy''=li 
x^-xy-\'y^  =  1 )  x^  +  xy-f  =  ^. 

6.  a;  — y  =  9    I  13.   a.-  +  y  =  a  I 
xy^8  =  0]                             4xy  -  a^  =  -  4^2  j 

7.  5:r  — 7y  =  0  ^      14.   :r  — y  =  l    ^j 

5^_l?£^  =  4-7y^  ^      y^^i 

4  J  y      :r  J 


SIMULTANEOUS    QUADRATIC    EQUATIONS. 


225 


15.   x^-{-9x7/  =  S4cO) 


a      i 


26.    x^  —  7/^  =  a^ 
x-y 

16.  a;  +  y  =  6      |  Vin.   x^  —  xy  =  a^^h'^ 

17.  3a:?/  +  2a;  +  y  =  485\       28.   x^  —  f  =  ^ab\ 


3a;-2y  =  0 


18.  x  —  y^ 

19.  a.-3  + 7/3  =  2728 
^-rry  +  y'  =  124: 

2D.    a;  +  2/  =  a     1 
=  5M 


xy  =  a^ 
29.    :ry  =  0 


5«   / 
■c} 


x^  +  f 


21.   :r2-y2  =  0 
3:r2-4 

0.'  +  ?/        X 


22. 


3a;^  — 4a:?/-f-5?/^ 

y^ 

x  —  y      x-\-y 
^  +  2/2  =  45 

1  .  1 


^o} 


30.  :f?-\-xy-\-if  =  Z'J 
.'c^  +  a:y  +  y^  =  481 

31.  ar^  =  aa:  -f  iy  ) 

y^  =  ay  +  Ja;  J 

32.  a:-y-2  =  0  1 
15(a;=^-y2)  =  16a:yJ 


10 


33. 


:r  +  y  a;  — y_89 
a;  —  y  a;  4-  y  40 
6a;  =  20y  +  9 


23.    _  +  _  =  5 
X     y 

1       , 


17 


x-\-\      y+1      12 
24.    :z;2_-^2/_|_y2^7         | 
/'»=133J 


34.    ^  +  ^=1 

X      y 


x'^r^f^-y^ 


35.    .T2  +  y2=7  +  :ry    | 
6  a:y  —  1  i 


5^-3 +  y3 

25.   a:  +  y  =  4      1  36.    a;^  —  y*  =  3093 ") 

^^  +  y*  =  82J  a:-y  =  3  J 

37.  |(:r-l)-|Cr  +  l)(y-l)  =  -lU 

i(y  +  2)-i(:i-  +  2)  J 

38.  105."2  +  15a'y  =  3a5-2an 
10y2+15a:y  =  3a6-252j 


226  ALGEBRA. 


Exercise  XCV. 

1.  If  the  length  and  breadth  of  a  rectangle  were  each  in- 

creased by  1,  the  area  would  be  48  ;  if  they  were  each 
diminished  by  1,  the  area  would  be  24.  Find  the 
length  and  breadth.     ^^  ^ 

2.  The  sum  of  the  squares  of  the  two  digits  of  a  number  is 

25,  and  the  product  of  the  digits  is  12.  Find  the 
number. 

3.  The  sum,  the  product,  and  the  difference  of  the  squares, 

of  two  numbers  are  all  equal.     Find  the  numbers. 
Note.    Represent  the  numbers  hj  x  +  y  and  x  —  y,  respectively. 

4.  The  difference  of  two  numbers  is  f  of  the  greater,  and 

the  sum  of  their  squares  is  356.  What  are  the  num- 
bers ?    /  (^      i  0 

5.  The  numerator  and  denominator  of  one  fraction  are  each 

greater  by  1  than  those  of  another,  and  the  sum  of 
the  two  fractions  is  1^^ ;  if  the  numerators  were  in- 
terchanged the  sum  of  the  fractions  would  be  1^. 
Find  the  fractions. 

6.  A  man  starts  from  the  foot  of  a  mountain  to  walk  to  its 

summit.  His  rate  of  walking  during  the  second  half 
of  the  distance  is  \  mile  per  hour  less  than  his  rate 
during  the  first  half,  and  he  reaches  the  summit  in 
5  J  hours.  He  descends  in  3|  hours,  by  walking  1 
mile  more  per  hour  than  during  the  first  half  of  the 
ascent.  Find  the  distance  to  the  top  and  the  rates 
of  walking. 
Note.  Let  1x=  the  distance,  and  y  miles  per  hour  =  the  rate  at  first. 
Then*  -  +  -^—  =  5^  hours,  and  -^  =  3f  hours. 

y   y-h  y+i 


SIMULTANEOUS    QUADRATIC    EQUATIONS.  227 

7.  The  sum  of  two  numbers  which  are  formed  by  the  same 
two  digits  in  reverse  order  'is  -f  f  of  their  difference ; 
and  the  difference  of  the  squares  of  the  numbers  is 
3960.     Determine  the  numbers. 

^  3    -  ^  The  hypotenuse  of  a  right  triangle  is  20,  and  the  area 
'if-  -5-  <?  L        of  the  triangle  is  96.     Determine  the  sides.      jX    "  ^  ^ 
^  Note.   The  square  on  the  hypotenuse  =  sum  of  the  squares  on  the 

aides ;  and  the  area  of  a  right  triangle  =  \  product  of  sides. 

9.  Two  boys  run  in  opposite  directions  round  a  rectangular 
field  the  area  of  which  is  an  acre ;  they  start  from 
one  corner  and  meet  13  yards  from  the  opposite  cor- 
ner ;  and  the  rate  of  one  is  |-  of  the  rate  of  the  other. 
Determine  the  dimensions  of  the  field. 

10.  A,  in  running  a  race  with  B,  to  a  post  and  back,  met 

him  10  yards  from  the  post.  To  make  it  a  dead  heat, 
B  must  have  increased  his  rate  from  this  point  41-^ 
yards  per  minute  ;  and  if,  without  changing  his  pace, 
he  had  turned  back  on  meeting  A,  he  would  have 
come  4  seconds  after  him.  .How  far  was  it  to  the 
post? 

11.  The  fore  wheel  of  a  carriage  turns  in  a  mile  132  times 

more  than  the  hind  wheel ;  but  if  the  circumferences 
were  each  increased  by  2  feet,  it  would  turn  only  88 
times  more.     Find  the  circumference  of  each. 

12.  A  person  has  $6500,  which  he  divides  into  two  parts 

and  loans  at  diffei'ent  rates  of  interest,  so  that  the  two 
parts  produce  equal  returns.  If  the  first  part  had 
been  loaned  at  the  second  rate  of  interest,  it  would 
have  produced  $  180  ;  and  if  the  second  part  had  been 
loaned  at  the  first  rate  of  interest,  it  would  have  pro- 
duced $  245.     Find  the  rates  of  interest. 


CHAPTER    XVI. 

Simple  Indeterminate  Equations. 

242.  If  a  single  equation  be  given  which  contains  two 
unknown  quantities,  and  no  other  condition  be  imposed, 
the  number  of  its  solutions  is  unlimited;  for,  if  any  value  be 
assigned  to  one  of  the  unknown  quantities,  a  corresponding 
value  may  be  found  for  the  other.  Such  an  equation  is 
said  to  be  indeterminate. 

243.  The  values  of  the  unknown  quantities  in  an  inde- 
terminate equation  are  dependent  upon  each  other;  so  that, 
though  they  are  unlimited  in  number,  they  are  confined  to 
&  particular  range. 

This  range  may  be  still  further  limited  by  requiring 
these  values  to  satisfy  some  given  condition ;  as,  for  instance, 
that  they  shall  be  positive  integers. 

244.  The  method  of  solving  an  indeterminate  equatiou 
in  positive  integers  is  as  follows : 

(1)    Solve  3:r  +  4^/  =  22,  in  positive  integers. 
Transpose,  3x  =  22  —  42/, 

the  quotient  being  written  as  a  mixed  expression. 

^  3 

Since  the  values  of  x  and  y  are  to  be  integral,  x  +  y  —  7  will  be 
integral,  and  hence,  ~  ^  will  be  integral,  though  written  in  the 
form  of  a  fraction. 

Let  ~  ^  =  m,  an  integer ; 


SIMPLE    INDETERMINATE    EQUATIONS.  229 

,     Then  1  —  y  =  3m, 

.•.?/  =  1  —  3m. 
Substitute  this  value  of  y  in  the  original  equation, 
3a;  +  4-  12m  =  22, 

.'.  a;  =  6  +  4ot. 
The  equation  y  =  1  —  3m  shows  that  m  in  respect  to  y  may  be  0, 
or  have  any  negative  value,  but  cannot  have  a  positive  value. 

The  equation  oj  =  6  +  4  m  shows  that  m  in  respect  to  x  may  be  0, 
but  cannot  have  a  negative  value  greater  than  1. 
.'.  m  may  be  0  or  —  1, 
and  then  a;  =  6,  3/  =  1, 

or  a;  =  2,  y  =  4, 

(2)    Solve  5a;  —  14?/  =  11,  in  positive  integers. 
Transpose,  5a;  =  11  +  142/, 

,  =  2  +  22/  +  !^. 

^  5 

.*.  — — — ^  must  be  integral. 
5 

Now,  if  — -— ^  be  put  =  w,  then  y  =    ^  ~    ,  a  fraction  in  form. 

5  4 

To   avoid   this  difficulty,   it  is  necessary  in   some  way  to  make 

the  coefficient  of  y  equal  to  unity.     Since  — — — ^   is   integral,    any 

multiple  of  — — — ^  is  integral.     Multiply,  then,  by  such  a  number  as 

5 
will  make  the  coefficient  of  y  greater  by  1  than  some  multiple  of  the 

denominator.     In  this  case,  multiply  by  4.     Then 

i+_16y  or  33/  +  ^-^  is  integral 
5  5 

.  4  +  y 
'*    5 

,'.  y  =•=  5m  —  4. 

Since       a;  =  -^  (11  +147/),  from  the  original  equation, 
.-.  X  =  14m  —  9. 

Here  it  is  obvious  that  m  may  have  any' positive  value,  and 

a;  =  5,  19,  33 

2/  =  1.  6,  11 


230  ALGEBRA. 

The  required  multiplier  can  always  be  found  when  the  coefficients 
are  prime  to  each  other,  and  it  is  best  to  divide  the  original  equation 
by  the  smaller  of  the  two  coefficients,  in  order  to  have  the  multiplier 
as  small  as  possible. 

245.  The  necessity  for  a  multiplier  may  often  be  obviated 
by  a  little  ingenuity.     Thus, 

The  equation         4  2/  =  29  —  7a;       may  be  put  in  the  form  of 
42/  =  29  -  8a;  +  x, 
.•.2/  =  7-2.r  +  ^, 
in  which  the  fraction  is  of  the  required  form. 

The  equation         bx  =  18  +  13y 
gives  a;  =  3  +  2t/  +  ^^"t^, 

in  which  -  ^^Jl  ig  of  the  required  form. 

246.  It  will  be  seen  from  (1)  and  (2)  that  wlien  only  pos- 
itive integers  are  required,  the  number  of  solutions  will  be 
limited  or  unlimited  according  as  the  sign  connecting  x  and 
y  is  positive  or  negative. 

(3)    Find  the  least  number  that  when  divided  by  14  and  5 
will  give  remainders  1  and  3  respectively. 

If  N  represent  the  number,  then 

N~\  aN-Z  '. 

=  X,  ami =  V. 

14  5  •^' 

.'.  N=  14a;  +  1,  and  N=  Sy  +  3, 
.-.  14a;  +  1  =  5y  +  3. 
5y  =  14a;-2, 
by  =  Ibx  —  2  —  X, 

o        2  +  X 

.-.  V  =  2x — . 

^  5 

2  +  X 
Let  — — —  =  m,  an  integer  ; 

5 
.-.  a;  =  5m  -  2. 

y  =  \{l-i:X  —  2),  from  original  equation, 
.*.  y  =  14m  —  6. 
If  m  =  1,  a;  =  3,  and  y  =  8, 

.-.  iV-  14.T  +  1  =  57/  +  3  =  43.  Ans. 


SIMPLE   INDETERMINATE   EQUATIONS.  231 

(4)    Solve  5  a;  +  6y  =  30,  so  that  x  may  be  a  multiple  of  y, 
and  both  positive. 

Let  X  =  my.  • 

Then  (5m  +  6)y  =  30, 

30 


y  = 


and 


5m  +  6 
30  m 

5m  +  6 


Ifm  =  2,  .r-  3f,  y  =  1|. 

If  m  =  3,  a;  =  4f ,  y  =  If 

(5)    Solve  14a; -f  22?/  =  71,  in  positive  integers. 

^  14 

If  we  multiply  the  fraction  by  7  and  reduce, 

the  result  is  —  4y  +  i, 

a  form  which  shows  that  there  can  be  no  integral  solution. 

There  can  be  no  integral  solution  oiax  ±hy  =  c\i  a  and  b  have  a 
common  factor  not  common  also  to  c ;  for,  if  c?  be  a  factor  of  a  and 
also  of  h,  but  not  of  c,  the  equation  may  be  written, 
nidx  ±  ndy  =  c, 
or  mx  db  ny  =  — ,  a  fraction. 


Exercise  XCVI. 
Solve  in  positive  integers : 

1.  2a:  +  lly  =  49.  5.  3a;  +  8y  =  61. 

2.  7a;  +  3y  =  40.  6.  8a;  +  5y  =  97. 

3.  5a:-f7y  =  53.  7.  16a:  +  7y  =  110. 

4.  a;+10y  =  29.  8.  7a; +10?/ =  206. 
Solve  in  least  positive  integers : 

9.    12a:-7y=l.  12.  23ar-9?/  =  929. 

10.  5a; -17?/ --23.  13.  23.r-33y  =  43. 

11.  23y-13a;  =  3.  14.  555a;  -  22y  =  73. 


232  ALGEBRA. 

15.  How  many  fractions  are  there  with  denominators  12 

and  18  whose  sum  is  -|-|-  ? 

16.  "What  is  the  least  number  which,  when  divided  by  3 

and  5,  leaves  remainders  2  and  3  respectively  ? 

17.  A  person  counting  a  basket  of  eggs,  which  he  knows 

are  between  60  and  60,  finds  that  when  he  counts 
them  3  at  a  time  there  are  2  over ;  but  when  he 
counts  them  5  at  a  time  there  are  4  over.  How 
many  are  there  in  all  ? 

18.  A  person  bought  40  animals,  consisting  of  pigs,  geese, 

and  chickens,  for  §40.  The  pigs  cost  §5  a  piece,  the 
geese  $1,  and  the  chickens  25  cents  each.  Find  the 
number  he  bought  of  each. 

19.  Find  the  least  multiple  of  7  which,  when  divided  by  2, 

3,  4,  6,  6,  leaves  in  each  case  1  for  a  remainder. 

20.  In  how  many  ways  may  100  be  divided  into  two  parts, 

one  of  which  shall  be  a  multiple  of  7  and  the  other 
of  9? 

21.  Solve  18:r  —  5?/  =  70  so  that  y  may  be  a  multiple  of  x^ 

and  both  positive. 

22.  Solve  8a7+  12y  =  23  so  that  x  and  y  may  be  positive, 

and  their  sum  an  integer. 

23.  Divide  70  into  three  parts  which  shall  give  integral 

quotients  when  divided  by  6,  7,  8,  respectively,  and 
the  sum  of  the  quotients  shall  be  10. 

24.  Divide  200  into  three  parts  which  shall  give  integral 

quotients  when  divided  by  5,  7,  11,  respectively,  and 
the  sum  of  the  quotients  shall  be  20. 

25.  A  number  consisting  of  three  digits,  of  which  the  mid- 

dle one  is  4,  has  the  digits  in  the  units'  and  hundreds' 
places  interchanged  by  adding  792.  Find  the  number. 


SIMPLE   INDETERMINATE   EQUATIONS.  233 

26.  Some  men  earning  each  $2.50  a  day,  and  some  women 

earning  each  §1.75  a  day,  receive  altogether  for  their 
daily  wages  $44.75.  Determine  the  number  of  men 
and  the  number  of  women. 

27.  A  wishes  to  pay  B  a  debt  of  £1  12s.,  but  has  only  lialf- 

crowns  in  his  pocket,  while  B  has  only  4  penny-pieces. 
How  may  they  settle  the  matter  most  simply? 

28.  Show  that  323  :r  —  527y  =  1000  cannot  be  satisfied  by 

integral  values  of  cc  and  y. 

29.  A  farmer  buys  oxen,  sheep,  and  hens.     The  whole  num- 

ber bought  is  100,  and  the  whole  price  £100.  If  the 
oxen  cost  £5,  the  sheep  £1,  and  the  hens  Is.  each, 
how  many  of  each  did  he  buy  ? 

30.  A  number  of  lengths  3  feet,  5  feet,  and  8  feet  are  cut ; 

how  may  48  of  them  be  taken  so  as  to  measure  175 
feet  all  together  ? 

31.  A  field  containing  an  integral   number   of  acres  less 

than  10  is  divided  into  8  lots  of  one  size,  and  7  of 
4  times  that  size,  and  has  also  a  road  passing  through 
it  containing  1300  square  yards.  Find  the  size  of 
the  lots  in  square  yards. 

32.  Two  wheels  are  to  be  made,  the  circumference  of  one 

of  Avhich  is  to  be  a  multiple  of  the  other.  What  cir- 
cumferences may  be  taken  so  that  when  the  first  has 
gone  round  three  times  and  the  other  five,  the  differ- 
ence in  the  length  of  rope  coiled  on  them  may  be  17 
feet? 

33.  In  how  many  ways  can  a  person  pay  a  sum  of  £15  in 

half-crowns,  shillings,  and  sixpences,  so  that  the  num- 
ber of  shillings  and  sixpences  together  shall  be  equal 
to  the  number  of  half-crowns  ? 


CHAPTER    XVII. 

Inequalities. 

247.  Expressions  containing  any  given  letter  will  have 
their  values  changed  when  different  values  are  assigned  to 
that  letter ;  and  of  two  such  expressions,  one  may  be  for 
some  values  of  the  letter  larger  than  the  other,  for  other 
values  of  the  letter  smaller  than  the  other. 

Thus,  1  +  X  +  x^  will  be  greater  than  1  —  x  +  x^  for  all  positive 

values  of  x,  but  less  for  all  negative  values  of  x. 

«" 

248.  One  expression,  however,  may  be  so  related  to  an- 
other that,  whatever  values  may  be  given  to  the  letter,  it 
cannot  be  greater  than  the  other. 

Thus,  2x  cannot  be  greater  than  a^  +  1,  whatever  value  be  given 
to  X. 

249.  For  finding  whether  this  relation  holds  between 
two  expressions,  the  following  is  a  fundamental  proposition  : 

If  a  and  h  are  unequal,  a^  ^h^>2ah. 

For,  (a  —  hf  must  be  positive,  whatever  the  values  of  a  and  6, 

That  is,  (a  -  If  >  0. 

or  a'-2ah  ^VX)', 

:.  a^  +  ¥>2  ah. 

250.  The  princir)les  applied  to  the  solution  of  equations 
may  be  applied  to  inequalities,  except  that  if  each  side  of 
an  equality  have  its  sign  changed,  the  inequality  will  be 
reversed. 

Thus,  if  a  >  6,  then  -  a  will  be  <  -  6. 


INEQUALITIES.  235 


(1)  If  a  and  h  be  positive,  show  that  c^  +  5^  is  >  c^h  +  aV^. 

a?  -\-lJ^>  a%  +  ah^, 
if  (dividing  each  side  by  a  +  h\ 

a"  -ab  +  h^>  ab, 
if  a«  +  6*  >  2  ab. 

But  a*  +  6M8>2a6,  §249. 

.-.  a^  +  P>  a'b  +  ab*. 

(2)  Sho^Y  thsita^  +  b^ -}-(^  is  >ab-^ac'}- be. 

Now,  o'  +  &*'  is  >  2  ab, 

a*  +  c*  is  >  2  ac,  §  249. 

6*  +  cMs  >  2  be. 
By  adding,  2a«  +  262  +  2c2  is  >  2a6  +  2ac  +  26c, 
.-.  a'^  +  i'^  +  c^  is  >  a6  +  ac  +  be. 


Exercise  XCVII. 
Show  that,  the  letters  being  unequal  and  positive : 

1.  a^  +  db^is>2b(a  +  b).  2.    a^b  + aP  is>2a^b^ 

3.  (a^  +  P)  (a*  +  b*)  is  >  (a^  +  ^3)2. 

4.  a^^  +  aJ'c  +  aZ^2  _|.  p^  _^  ^^2  ^  5^.2  ^^  >  q  ^ j^ 

5.  The  sum  of  any  fraction  and  its  reciprocal  is  >  2. 

6.  If  :r^=a^-{-^2,  and  7/^  =  (^-}-d^,xi/is  >ac-]-bd,  or  ad-\-bc. 

7.  a^>  +  «c  +  bc<(a  +  6  -  c/  +  (a  +  c  -  Z>/+(Z^  +  c  -  a/. 

8.  Which  is  the  greater,  (a^  +  P)  (c^  +  cP)  or  («(?  +  Z>c^)2  ? 

9.  Which  is  the  greater,  tti^ -}- m  or  rn^ -{- 1  ? 

10.  Which  is  the  greater,  a*  —  Z>*  or  4  a^  (a — i)  ^Yhen  a  h  >  5  ? 

11.  Which  is  the  greater,  a/t"  +  -V  ~  or  Va  +  V^? 

12.  Which  is  the  greater,  or  -^^ — -  ? 

13.  Which  is  the  greater,  71  +  -^  oi'  7  +    " 


CHAPTER  XVIIL 
Theory  of  Exponents. 

251.  The  expression  a**,  when  n  is  a  positive  integer,  has 
been  defined  as  the  product  of  n  equal  factors  each  equal 
to  a.  §24. 

And  it  has  been  shown  that  a*"  X  a"  =  «*"+".  §  66. 

That  ar-~al'  =  a*""",  if  m  be  greater  than  n ;  §  93. 

or ,  if  m  be  less  than  n.  §  94. 

And  that  {cTY  =  a'^^  §  199. 

Also,  it  is  true  that  a*'xh''  =  (ahy ;  for 
(ahy  =  ah  taken  n  times  as  a  factor, 

=  a  taken  n  times  as  a  factor  X  b  taken  n  times  as  a 
factor  =  a"  X  S". 

252.  Likewise,  Va,  when  n  is  a  positive  integer,  has  been 
defined  as  one  of  the  n  equal  factors  of  a  (§  203)  ;  so  that  if 
"y/a  be  taken  n  times  as  a  factor,  the  resulting  product  is  a ; 
that  is,  K'y/aJ  =  a. 

Again,  the  expression  Va**  means  that  a  is  to  be  raised 
to  the  Tnth  power,  and  the  nth  root  of  the  result  obtained. 

And  the  expression  (Va)"*  means  that  the  nth  root  of  a 
is  to  be  taken,  and  the  result  raised  to  the  mth  power. 

It  will  thus  be  seen  that  any  proposition  relating  to  roots 
and  powers  may  be  expressed  by  this  method  of  notation. 
It  is,  however,  found  convenient  to  adopt  another  method  of 
notation,  in  which  fractional  and  negative  exponents  are 
used. 


THEORY    OF    EXPONENTS.  237 

253.  The  meaning  of  a  fractional  exponent  is  at  once  sug- 
o-ested,  by  observing  that  the  division  of  an  exponent,  when 
the  resulting  quotient  is  integral,  is  equivalent  to  e:^tracting 
a  root.  Thus,  a^  is  the  square  root  of  a®,  and  3,  the  expo- 
nent of  a^,  is  obtained  by  dividing  the  exponent  of  a^  by  2. 

If  this  division  be  indicated  only,  the  square  root  of  a^ 
will  be  denoted  by  a^,  in  which  the  denominator  denotes 
the  root,  and  the  numerator  the  power.  If  the  same  mean- 
ing be  given  to  an  exponent  when  the  division  does  not  give 
an  integral  quotient,  a^  will  represent  the  square  root  of 
the  cube  of  a ;  and,  in  general,  a«,  the  nth  root  of  the  mth 
power  of  a.  This,  then,  is  the  meaning  that  will  be  as- 
signeid  to  a  fractional  exponent,  so  that  in  a  fractional 
exponent 

254.  The  numerator  will  indicate  a  power,  and  the  de- 
nominator a  root, 

255.  The  meaning  of  a  negative  exponent  is  suggested 
by  observing  that  in  a  series  of  descending  powers  of  a, 

a* a*,  a*,  'c?,  a^,  a\ 

the  subtraction  of  1  from  the  exponent  is  equivalent  to  di- 
viding by  a ;  and  if  the  operation  be  continued,  the  result 
is. 


Then, 


This,  then,  is  the  meaning  that  will  be  assigned  to  a  neg- 
ative exponent,  so  that, 

256.  A  number  luith  a  negative  exponent  will  denote  the 
reciprocal  of  the  number  with  the  corresponding  positiva 
exponent. 


(^', 

a'\ 

a~ 

2 

a~ 

3 

a'\ 

....  ( 

a" 

a«: 

_a  _ 
a 

^l 

; 

a~ 

l_, 

=  1- 

a  — 

.1 

a 

r': 

_1  . 

-a 



1 
'a' 

a-"  = 

_± 

a 

a" 

238  ALGEBRA. 


It  may  be  easily  shown  that  the  laws  which  apply  to  pos- 
itive integral  exponents  apply  also  to  fractional  and  negative 
exponents. 

257.    To  show  that  c^  xh^  =-  (abf  : 
al  X  &  »  =  Va^  X  Vb^, 

=  (a5)»»  (by  definition). 

Likewise  a    X  Z>"  X  c"  =  (abc)^,  and  so  on. 


258.    To  show  that  (a^)n=a^: 


Let 

a;  =  (a-)^ 

Then 

x^  =  a"»,  and  x""*  =  a. 

.:  X  =  a^^. 

But 

X  =  (a"*)'"*          (by  supposition), 

259.  To  show  that  a"*  X  a""  =  a"*"" : 
No-w:  a"»  X  a-"  =  a*"  X  — , 

-»=  —  =  a"*-'*  if  m  >  n  •  ^  93. 

a" 

or  =— ^ifm<n,  §94. 

==  a'^^""*)         (by  definition), 

260.  In  like  manner' the  same  laws  may  be  shown  to 
apply  in  every  case. 


THEORY    OF    EXPONENTS.  239 

261.    Hence,  whether  m  and  n  be  integral  ov  fractional, 
positive  or  negative  : 

I.  a"'Xa^  =  a'"+".  III.  («"•)«  =  a'"". 

IT.  a"*  -^  a"  =  a*"-".  IV.  a"*  X  ^"^  =  (aJ)"*. 

Exercise  XOVIII. 
Express  with  fractional  exponents : 

2.  -v/V^^;    ■v'^y?;   "^/^M/;  5Va^^^*.  ^ 
Express  with  radical  signs : 

3.  a^;  a^h^]  4:X^y'^]    Zx^y-^.  ' 
Express  with  positive  exponents  : 


3-1 6^-'" 

Write  in  the  form  of  integral  expressions : 
^    3a;?/       z    ^    a  ^      <^    .  ^~^  .   ^~^ 

Simplify : 

6.  o^  X  a^  ;  Z»^  X  Z>^  ;   c^  X  cT2  ;  d^  x  (fT^. 

7.  m^  X  m~^  ;  n^  x  n'Ti ;  a^  x  a^  ;  a^  X  a~i 

8.  a^  X  Va  ;   c"  ^  X  V^;  y^  X  -v^y  ;  :r^  X  V^\ 

9.  aS^cXa'Hcs;  a^Z^^c"^  X  an"^<?^6?. 

10.   x^  y 5  2^  X  a;~  3  y~  ^'  2"  ^  ;  x^  y ^  ^  X  a;~  ^  y~  ^2"  i 


240  ALGEBRA. 


12.  a*-f-a^;  c^ --- a"  ■  n'^^vA;  a^-^Va". 

13.  (a«)^H-(a°)^;  (e"^)';   (m"^)'*;   {n^'^  ■   (a:^)! 

14.  (p-^)-^;  ijpy^]  {x-^y^y^;  (a*Xa')-T3. 
15".  (4a- 3)- 1;  (Tih-^y^;  (64c^«)-*;  (32 O^. 

16  A^^'V'-    ^_^^V'.    (^^a-^Y^^-    (2-56)- 4 


262.  The  laws  that  apply  to  the  exponents  of  simple 
expressions  also  apply  to  the  exponents  of  compound  ex- 
pressions. 

(1)  Multiply  3/4  +  y^  +  y^  +  1  by  ?/4  _  l. 

2/3  +  3/^  +  2/1  +  1 

3/^-1 

y  +yl  +  yh  +  y\ 

-yl-yl-y\-l 
y  —  1  2/  —  1-  '^^^■ 

(2)  Divide  x^  +  a:^  -  12  by  :ri  -  3. 

{x^  -  3 
g  ,  ^x'f  +  4 

4a;^  -  12 

4x^  -  12  2;i  +  4.  ^.'?s 

Exercise  XCIX. 


Multiply  : 

2.  :r""*~''  —  y**  by  x"  -f  t/"'"-". 

3.  a:*-2:r3  +  l  by.-r^-l. 

4.  8a^  +  4a757-L5aH^  +  9JHy2a^-/.f 


THEORY    OF    EXPOXENTo.  241 

6.  cH-^  +  2  +  a'^V  by  aH'''  -  2  -  a-H\ 

7.  4:x-^ -\-^x-^  +  2x-^  +  \  by  x-^ - x-"^  -f-  1. 

Divide  : 
a   x^""  —  y'*'*  by  a;"  —  y". 
9.   rt;  +  y  +  z  — Sajiy^z^  by  a;3 +yi-|-2;3. 

10.  x-\-y  by  a;"5  —  a;5y3-|~^^3/^  —  x^y^  +  2/*- 

11.  x^y~-  +  2  -f-  x~^y'^  by  a;?/"^  -{-  x''^y. 

12.  a-*  +  a-2  6--  +  ^-^  by  a-2-a-i^*-i  +  5-3. 

Find  the  squares  of: 

13.  4a^-^  a^-h^]  a  +  a'^;  2aU^  — a-U^. 

If  a  =  4,  5  =  2,  c  =  1,  find  the  values  of: 

14.  aU;  bab-"^]  2{ab)\ ;  ar^b'^c^-  l^a^H^^ 

15.  Expand  (ai-5i)3;  (2a7-i  +  :r)^  (ab-' -  by'f. 

Extract  the  square  root  of: 

16.  9^-*  -  18x-^yi  +  15x-^y  -  6x-'^yi  +  y". 

Extract  the  cube  root  of : 

17.  Sa^  +  Ux"  -SOx-So-i-  4:bx-'^  +  27x-^  ~  21x-\ 

Resolve  into  prime  factors  with  fractional  exponents  : 

18.  -v/ll,   </72,   -v/96,   -v/el ;  and  find  their  product. 

Simplify : 

19.  K^'")'x(.r^^)-2j3^2.        20.    (x^^'' X  x-'^)sh2, 

21.  3(a^  -h  5^)2-4  (ai  +  i^)  (J  -  bl)  +  (a^-  2Z>^f. 

22.  J(a'»)'"-^.|'^.  24.    [J(a-'")-"p]»--[J(a'")"|-^]-'. 

23.  f^^^ Wf-i—l       .      25.    '^^^ ^^ . 


242  ALGEBRA. 


Eadical  Expressions. 

263.  An  indicated  root  that  cannot  be  exactly  obtained 
is  called  a  surd,  or  irrational  number.  An  indicated  root  that 
can  be  exactly  obtained  is  said  to  have  iheforTii  of  a  surd. 

264.  The.  required  root  shows  the  order  of  a  surd;  and 
surds  are  named  quadratic,  cubic,  biquadratic,  according  as 
the  second,  third,  ot  fourth  roots  are  required. 

265.  The  product  of  a  rational  factor  and  a  surd  factor  is 
called  a  laixed  surd;  as,  3 V2,  &  Va. 

266.  When  there  is  no  rational  factor  outside  of  the  rad- 
ical sign,  the  surd  is  said  to  be  entire;  as,  V2,  Va. 

267.  Since  Va  X  V5  X  V^=  ^abc,  the  product  of  two 
or  more  surds  of  the  same  order  will  be  a  radical  expression 
of  the  same  order  consisting  of  the  product  of  the  numbers 
under  the  radical  signs. 

268.  In  like  manner,  Va^  =  Vc?  X  V^=  a  VJ.    That  is, 
A  factor  under  the  radical  sign  whose  root  can  be  taken, 

may,  by  having  the  root  taken,  be  removed  from  under  the 
radical  sign. 

269.  Conversely,  since  aV^=  Vo^, 

A  factor  outside  the  radical  sign  unay  be  raised  to  the  cor- 
responding power  and  placed  under  it. 


•^i=^R=^ 


7  Va 


a         \ab  I   7  V,  1       1     /-7 


<r<%-<^-i=i^^ 


RADICAL    EXPRESSIONS. 


243 


270.  A  surd  is  in  its  simplest  jona  when  the  expression 
under  the  radical  sign  is  integral  and  as  small  as  possible. 

271.  Surds  which,  when  reduced  to  the  simplest  form, 
have  the  same  surd  factor,  are  said  to  be  similar. 


IT         4  ["5^ 
\12'      \26V 


Simplify : 
V50;     -^108;     Vl^' ] 

(1)  V50=  V2Kx2  =  bV2. 

(2)  vTjOS  =  V2nxl  =  3v^4. 

(3)  Vt^''  =  y/l^fxf  =  y^l^. 


V296352. 


,Kx     A  5a         4/40ak-»        1    ^, — 

(6)    v/296352. 


23 

296352 

22 

37044 

32 

9261 

3 

1029 

7 

343 

7 

49 

Hence,  296352  =  2^  x  3^  x  7=^. 

.-.  ^^/296352  =  y/f"  x  \^=*  X  v^l^, 
=  7x3x2^', 
=  42  \^.  An^. 


In  simplifying  numerical  expressions  under  the  radical  sign,  tli« 
method  employed  in  (6)  may  be  used  with  advantage  when  the  factor 
whose  root  can  be  taken  is  not  readily  determined  by  inspection. 

Exercise  C. 
Express  as  entire  surds  : 

1.  3V5;  3V2I;  5V32;  a%-Jhc;  x^s/¥f. 

2.  3?/^V:^y;  '2x-\fxy;  a^-VoFl?;  Se^^abc;  babc^abc'^r 


3.  f 


lOVftf;  (x  +  y)^ 


SL 


:>?  +  2xy  +  f 


244  ALGEBRA. 

Express  as  mixed  surds  : 

4.  VP3A;   VS^;   ■^54aV?/3;   V24 ;   Vl25^^. 

5.  -^v/IOOO^;   -\/imxy;   </lOSm^n'' ;   ^Z  1^12 a''b'\ 

6.  ^a*-3a«6  +  3a^62-a^3;   VSOa^- 100 a6  + 50^1 

Simplify : 

7.  2</80^¥?;  7V'396^;  9^8l^yi;  5V726. 

8.  VI;  Vlii;  V3i;  fV90|;  2VJ. 

3  l2xy^  .    3/—  .  a  4  /  6  ISo^ 

10.    -^:  —2—:   ^^^'^^-tV:   ^^^l^f!^'^^ 


V5'  V1701'  V  2'  yv^yy  ■'  v  t'VV « 

11.  (a:r)  X  (i'a;)^  (2  a'b')  X  (b'a^)^ ;  5(3a'b'y)x(a'b-U/)K 

12.  Show  that  V20,  V45,  Vf  are  similar  surds. 

13.  Show  that  2-v/f7P,  </8I?,  i^j  are  similar  surds. 

14.  If   V2  =  1.414213,  find  the  values  of 

V50;   fV288;    — ^ 


V2'    V450 

272.    Surds  of  the  same  order  may  be  compared  by  ex- 
pressing them  as  entire  surds. 

Ex.  Compare  |V7  and  fVlO. 

|v'7   =V^, 
fVIo  =  V^. 

V^  =  V^-,  and  VV-  =  ^^• 
As  V-^  is  greater  than  V-W",  1^1^  is  greater  than  |  Vj. 


RADICAL    EXPRESSIONS.  245 

273.  The  product  or  quotient  of  two  surds  of  the  same 
order  may  be  obtained  by  taking  the  product  or  quotient 
of  the  rational  factors  and  the  surd  factors  separately. 

(1)  2  V5  X  5  V7  =  10  V35. 

(2)  9V5--3V7  =  3V|  =  3V||  =  fV35. 

Exercise  CI. 

1.  Which  is  the  greater  SVT  or  2Vl5? 

2.  Arrange  in  order  of  magnitude  9  VS,  6  V7,  5  VlO. 

3.  Arrange  in  order  of  magnitude  4  V4,  3V5,  5V3. 

4.  Multiply  3 V2  by  4V6  ;  f  VIO  by  ^VIS. 

5.  Multiply  5  Vf  by  f  Vl62  ;  1^4  by  2^2. 

6.  Divide  2V5  by  3 Vl5  ;  f  V2l  by  ^W^. 

7.  Simplify  f  V3  X  |  V5  ~  f  V2. 

„    o.      y,    2Vl0^7V48      4VT5 

8.  Simplify zn  X 1=  -. 1^. 

8V27     5Vl4      15V21 

9.  Simplify  2^/4x5^32 --\/i08. 

274.  The  order  of  a  surd  may  be  changed  by  changing 
the  power  of  the  expression  under  the  radical  sign.      Thus, 

V5=^25;   ^=-^?. 
Conversely,       V25  =  Vo  ;   -\c^  =  Vc  ; 
or,  in  general,  "Vc"  =  V^. 


&"■ 


In  this  way,  surds  of  different  orders  may  be  reduced  to 
the  same  orden\  and  may  then  be  compared,  multiplied,  or 
divided. 


246 


ALGEBRA. 


(1)  To  compare  V2  and  VS. 

^  =  3^  =  3B=^2=^. 
.-.  VS  is  greater  than  V^. 

(2)  To  multiply  V^  by  V6a;. 

.^i^  =  (4a)^  =  (4a)«  =  v^(4^2  _  ^/Xeo^ ; 
V6^  =  (6ar)i  =  (6a:)i  =  \^{6^  =  v^2l6^. 

.-.  V^  X  VQ~x  =  VlQ^^  X  </2m?, 
_^  =  y/l6a^  X  216x3, 

=  v'24a2  X  23  X  3^"?, 
=  v^26  X  2  X  S^'aH^, 
=  2\/54a^.  Tins. 


^3)    To  divide  -v^  by  V63. 

.^3^  =  (3a)5  =  (3a)i  =  v^^^  =  ^C^^*; 
VG6  =  (66)^  =  (66)i  =  ^&^pf  =  ^e/2l6F. 


21663 


\2463       \ 


23x3^3' 


6/23x3^0-^63 


"WxV^^  =th^^^^^^^'-  ^^^^ 


Exercise  CII. 
Arrange  in  order  of  magnitude  : 

1.  2S^3,  3V2,  1^4.  3.   2^22,  3-\/7,  4V2. 

2.  Vf,  -v/||.  4.   3Vi9,  5a/2,  3^3. 


EADICAL    EXPRESSIONS.  247 

Simplify : 

5.  2 Va^  X  -^SM  X  V2^  ;   ^/a^xif  x  V^. 

6.  S(4:aPfs  ~  (2a%)i  ;  (2a%')^  X  {a'b')^  -  (a«5«)l 

7.  (2  ab)i  X  (3  ab'fs  -f-  (5  a^^)^  ;  4 Vl2  -f-  2  VS. 

9.   (7V2-5V6-3V8  +  4V20)x3V2. 

10.  V(if7  X  V(MT ;  ^v^(4^"  X  ■</(2^. 

11.  (^y^)^x(^/^^)*;  aU-^ctcTt-a^Z^-Ac-Hc^H 

275.  In  the  addition  or  subtraction  of  surds,  each  surd 
must  be  reduced  to  its  simplest  form  ;  and,  if  the  resulting 
surds  be  similar, 

Add  the  rational  factors,  and  to  their  sum  annex  the  com- 
mon surd  factor. 

If  the  resulting  surds  be  not  similar, 
Connect  them  with  their  proper  signs. 

276.  Operations  with  surds  will  be  more  easily  performed 
if  the  arithmetical  numbers  contained  in  the  surds  be  ex- 
pressed in  their  prime  factors,  and  if  fractional  exponents 
be  used  instead  of  radical  signs. 

(1)    Simplify  V27  +  V48  +  Vl47. 

V27==(3^)^=3x35  =  3\/3; 
Vis  =  (2^  X  3)^  =  22  X  3^  =  4  X  si  =  4\/3  ; 
Vl47  =  (7*  X  3)^  =  7  X  3'  =  7  Vs. 
.-.  V27  +  V48  +  Vl47  =  (3  +  4  +  7)  V3  =  14  Vs.  ^ns. 


248 


ALGEBRA. 


(2)  Simplify  2\/3i^- 3-^/40. 

2\/320  =  2  (26  X  5)^  =  2  X  2^  X  5'  =  8  v^; 
3  v^  =  3  (2=^  X  5)5  =  3  X  2  X  5^  =  6v^. 
.^.2</^-^^/40  =  8\^-G^/5  =  2^/5.  Ans. 

(3)  Find  the  square  root  of  Vsl. 

The  square  root  of  v^  =  (Sl^j^  =  SI*  =  (3*)* 
=  3i  =  (32)l=^9. 

(4)  Find  the  cube  of  |-^. 

The  cube  of  |^=  {^f  x  (2«/  =  i  x  2^  =  iV2. 

Exercise  CIII. 
Simplify : 

1.  V27  +  2V48  +  3VI08;  3V1000  +  4V50  + 12V288 

2.  's/l2S+-\/Q8Q-\-^/l6;  7^54  +  3^16+ "v^iB^. 

3.  12V7i2-3Vl28;  7-^-3-^/1029. 

4.  2V3  +  3Vl|-V5l;  2V|+V60-VT5-V|. 

6.  V4^+ V25^^-(a-55)V^. 

7.  cV^iFF?  ~  a</^'7^  +  bV^^\ 

a   2^/40 +  3^/l08+ ^500 -■v/320- 2^^1372. 


10.      (g^fg)";      (V27)i 


9.  (2-Vs^f;  (3-v/3)2. 

11.  (^81)4;  (^/512)^  (^256)-i;   ^16;  a/2T 

12.  ^4;   ^3^;   ^32;   ^^243;   </m ;   ^Z^^. 

13.  a/8^«;   -v/9^*;   ^16^^;  -v^Sl^^. 


RADIt3AL   EXPRESSIONS.  249 


14.  (-s/sy-  (-^27)^  (V64)«;  (V4)l 

15.  (a</a)-';  (x-\/xy^;  (/Vp)^;  (a-^^O"^. 


Expand  by  the  method  explained  in  §  201 : 

16.  ( Va  +  Vbf ;  {</m^  +  Vl^f  ;  (  V^  -  ^Vbf. 

17.  (2a2-iVa)«;  (2V^^-\,ff;   (^  -  ^^ . 


19.        r 


20. 


Find  the  square  root  of : 

21.  a;^"*  +  6:i;^**y"+ll:r'*?/2»»-f  6a;'"2/''*  +  2/^r 

22.  1  +  4:r-i  -  22;-!  -  4^'"*  +  25:^-^  -  24:^'^  +  IQx'^. 

o 

277.  If  we  wish  to  find  the  approximate  value  of ,  it 

V2 
will  be  less  labor  to  multiply  first  both  numerator  and  de- 
nominator by  a  factor  that  will  render  the  denominator 
rational;  in  this  case  by  V2.     Thus, 

3    ^      3V2      _3V2 
V2      V2  X  V2        2    ' 

278.  It  is  easy  to  rationalize  the  denominator  of  a  frac- 
tion when  that  denominator  is  a  binomial  involving  only 
quadratic  surds.  The  factor  required  will  consist  of  the 
same  terms  as  the  given  denominator,  but  with  a  different 


250  ALGEBRA. 


sign  between  them.     Thus, —-f-  will  have  its  denomi- 

6-f-2V5 

nator  rationalized  by  multiplying  both  terms  of  the  fraction 
by  6  -  2 V5.     For, 

7  -  3  V5  __  (7  -  3  V5)  (6  -  2  V5) 
6  +  2  V5  ~  (6  +  2  V5)  (6  -  2  V5) 

„72-32V5^?_2V5. 


16 


279.  By  two  operations  the  denominator  of  a  fraction 
may  be  rationalized  when  that  denominator  consists  of  three 
quadratic  surds. 

Thus,  if  the  denominator  be  V6  +  V3  —  V2,  both  terms 
of  the  fraction  may  be  multiplied  by  V6  —  V3  +  V2.  The 
resulting  denominator  will  be  6  —  5  +  2 V6  =  1  +  2  V6  ; 
and  if  both  terms  of  the  resulting  fraction  be  multiplied 
by  1  —  2 V6,  the  denominator  will  become  1  —  24  =  —  23. 

Exercise  CIV. 

Find  equivalent  fractions  with  rational  denominators,  for 
the  following : 

1  3  7  4- V2.  6 

V7  +  V5  '  2  V5  -  V6  '  1  +  V2  '  5  -  2  Ve' 

^  a         ^     a-\-h    ^  2x—^xy 

Vb  —  Vc     a—-\Jh     ^xy  —  2y 

Find  the  approximate  values  of : 

3    _?_•  1  7V5         7  +  2VIQ 

V3'  V5-V2'  V7  +  V3'  7-2VIO 


radical  expressions.  251 

Imaginary  Expressions. 

280.  All  imaginary  square  roots  may  be  reduced  to  one 
form. 

V^  =  Va  X  (—  1)  =  a^  V^. 

281.  V—  1  means  an  expression  which,  when  multiplied 
by  itself,  produces  —  1.     Therefore, 

(V^f = ( v^)^  X  v^ = - 1  v^n: = -v^i ; 

( V^ri)^  =  ( viri)^  X  ( V^)»  -=  (- 1)  x  (- 1)  =  l ; 

and  so  on.     So  that  the  successive  powers  of  V—  1  form  the 
repeating  series,  +V— 1,  —  1,  —  V—  1,  +  1. 

(1)  Multiply  1  +  V^  by  1  —  V^. 

1  +  V=4  =  1  +  2 V^l ; 
1  _  VTl  =  1  _  2  V^l. 
(1  +  2  V=l)  (1  -  2  v^)  =  1  _  4  (-  1)  -  5. 

(2)  Divide  ^/—ab  by  V^^. 

and  V^b    =  b^V^l. 

V—ab      aHsV^ 


V^b         b^V^l 


V~a. 


Multiply:  Exercise  CV. 

1.  4  +  V~3by4-V^;  V3-2V^by V3>2V=^. 

2.  V54  by  V^  ;  aV^  by  xV^. 


252  ALGEBRA. 


3.  V—  a  +  V—  b  by  V^  —  V^;  a  V-  a-6*  by  V—  a^b^. 

4.  V=TO  by  V^  ;  2V3  -  6V^  by  4V3  -  V^. 
Divide  : 

5.  rrV—  1  by  yV—  1 ;  1  by  V—  1 ;  a  by  a^V—  1. 

6.  V=n:2  by  V^  ;  Vl5  by  V-b  ;   V=^  by  V=^20. 

Square  Root  of  a  Binomial  Surd. 

282.  The  product  or  quotient  of  two  dissimilar  quadratic 
surds  will  be  a  quadratic  surd.     Thus, 

■y/ab  X  -y/oFc  =  ab^c  ; 
Va6c  -~-  -y/ab  ~  Vc. 

For  every  quadratic  surd,  when  simplified,  will  have  under  the 
radical  sign  one  or  more  factors  raised  only  to  the  first  power ;  and 
two  surds  which  are  dissimilar  cannot  have  all  these  factors  alike. 

Hence,  their  product  or  quotient  will  have  at  least  one  factor 
raised  only  to  the  ^rsi  power,  and  will  therefore  be  a  surd. 

283.  The  sum  or  diffei^ence  of  two  dissimilar  quadratic 
surds  canrwt  be  a  rational  number,  nor  can  it  be  expressed 
as  a  single  surd. 

For  if  yfa  ±  yfh  could  equal  a  rational  number  c,  we  should  have, 
by  squaring, 

a  ±  2\/a6  +  6  =  c^; 
that  is.  ±  2Va6  =  c^  —  a  —  h. 

Now,  as  the  right  side  of  this  equation  is  rational,  the  left  side 
would  be  rational ;  but,  by  \  282,  y/ab  cannot  be  rational.  There- 
fore, Va  ±  Vi  cannot  be  rational. 

In  like  manner,  it  may  be  shown  that  Va  ±  y/h  cannot  be  ex- 
pressed as  a  single  surd  Vc. 


RADICAL    EXPRESSIONS.  253 

284.  A  quadratic  surd  cannot  equal  the  sum  of  a  rational 
aumber  and  a  surd. 

For,  if  Va  could  equal  c  +  V6,  we  should  have,  by  squaring, 
a  =  c^  +  2c\/6  +  6, 
and,  by  transposing,  2cV6^=  a  —  h  —  (?. 

That  is,  a  surd  equal  to  a  rational  number,  which  is  impossible. 

285.  If  a-\-  ^/b  =  x-\-  Vy,  then  a  will  equal  x  and  h 
will  equal  y. 

For,  by  transposing,  yTb  —  Vy  =  x  —  a\  and  if  h  were  not  equal 
to  ?/,  the  difference  of  two  unequal  surds  would  be  rational,  which  by 
\  283  is  impossible. 

.'.h  =  y  and  a  =  x. 

In  like  manner,  if  a  —  V6  =  x  —  Vy,  a  will  equal  x  and  h  will 
equal  y. 

286.  To  extract  the  square  root  of  a  binomial  surd  a  +  V^. 


Let  V  a  +  y/h  =  Vx  +  Vy. 

Squaring,  a  +  Vb  =  x  +  2Vxy  +  y. 

.'.  X  +  y  =  a,  and  2Vxy  =  Vb.  285. 

From  these  two  equations  the  values  of  x  and  y  may  be  found. 

This  method  may  be  shortened  by  observing  that,  since 
Vb  =  2Vx7/,  _ 

a  —  V^  =  x~  2-Vxy  -f-  y. 

By  taking  the  root,   v  «  _  Vb  =  V^  —  Vy. 

.-.  (Va+V3)  {-s/a-Vb)  =  ( V^  +  Vy)  ( V^  -  Vy). 
.*.  -\/c^  —  b  =^x  —  y. 
And,  as  a  —  a;  +  y, 

the  values  of  x  and  y  may  be  found  by  addition  and  sub- 
traction. 


264  ALGEBRA. 


(1)    Extract  the  square  root  of  7  +  4  VS. 

Let  V^JrVy=  V7  +  4  Vs. 

Then  Vx  -  Vy  =  V7  -  4  VS. 

By  multiplying,    x  —  y  =  V49  —  48, 

.•.a;-y  =  1. 
But  a;  +  2/  =  7, 

.'.  X  =  4,  and  3/  =  3. 
.-.  V^  +  Vy  =  2  +  Vs. 
.-.  V7  +  4  V3  =  2+  V3. 

Exercise  OVI. 
Extract  the  square  roots  of : 

1.  14  +  6V5.  6.   20-8V6.  11.  14-4V6. 

2.  17  +  4V15.         7.   9-6V2.  12.  38-I2VIO. 

3.  IO  +  2V2I.         8.   94-42V5.  13.  103  -  12 VTI. 

4.  I6  +  2V55.         9.    13-2V30.  14.  57-I2VI5. 

5.  9-2VI4.  10.    II-6V2.  15.  3i-VlO. 

16.  2a  +  2V^^'^=T2.         18.   87-12a/42, 

17.  o?-2h^a'-h\        19.    {a^bf-4:{a-h)Vab. 

287.  A  root  may  often  be  obtained  by  inspection.  For 
this  purpose,  write  the  given  expression  in  the  form  a-\-2-\/bj 
and  determine  what  two  numbers  have  their  sum  eq;ial  a, 
and  their  product  equal  h. 

(1)    Find  by  inspection  the  square  root  of  18  +  2  V77. 

It  is  required  to  find  two  numbers  whose  sum  is  18  anc?  whose 
product  is  77 ;  and  these  are  evidently  11  and  7. 
Then        18  +  2  Vfz  ==  11  +  7  +  2  VTHT?, 

=  (Vll  +  V7)2. 
That  is,     Vll  +  V7  =.  square  root  of  18  +  2  V77. 


RADICAL    EXPRESSIONS.  255 

(2)    Find  by  inspection  the  square  root  of  75  —  12-\/2l. 

It  is  necessary  that  the  coefficient  of  the  surd  be  2  ;  therefore, 
75  —  12  V2I  raust  be  put  in  the  form  of 

75  -  2  V756 
The  two  numbers  whose  sum  is  75  and  whose  product  is  756 
are  63  and  12. 

Then        75  -  2V756  =  63  +  12  -  2\/63  x  12, 

=  (V63  -  VUf. 
That  is,     \/63  -  >/l2  =  square  root  of  75  -  12  a/M; 
or,  3  V7  _  2V3  =  square  root  of  75  -  I2V2I. 

Equations  containing  Radicals. 

288.  An  equation  containing  a  single  radical  may  be 
solved  by  arranging  the  terms  so  as  to  have  the  radical  alone 
on  one  side,  and  then  raising  both  sides  to  a  power  corre- 
sponding to  the  order  of  the  radical. 


Ex.  VV^^-fa7  =  9. 


Vx*  _  9  =  9  -  a;. 
By  squaring,  ic*  —  9  =  81  —  18  a;  +  a*. 

18  a;  =  90, 
.•.  a;  ==  5. 

289.    If  two   radicals   be    involved,   two   steps   may    be 
necessary. 

Ex.  Va;  +  15  +  V^  =  15. 

Vx  +  15  +  \/a^=15. 
By  squaring, 

a;  +  15  +  2Va^  +  15a;  +  x  =  225. 


By  transposing,  2  vV+TSx  =  210  —  2x. 

By  dividing  by  2,         Va;' +  15  x  =  105  —  x. 
By  squaring,  ar'  +  15  x  =  11025  -  210  »  +  x*. 

225x  =  11025, 
.-.  X  =  49. 


256  ALGEBRA. 

Some  of  the  following  radical  equations  will  reduce  to 
simple  and  others  to  quadratic  equations. 


^  ,     ,                     Exercise  OVII. 

1.   ^x-^  =  2. 

6.    V:r+4+V2a;-l-=6. 

2.  2V3:r  +  4-a;==4. 

3.  3-V^-l  =  2:r. 

7.  V13:^:-1-V2^-1  =  5. 

8.  V4:  +  x+Vx  =  S. 

4.  ^Sx-2  =  2(x-4:).         9.    -V~2b  +  x+-V25-x-- 

5.  4aj-12V^-=16.  10.   x2  =  2l4- V^'-9. 

11.  20^-^/8^  +  26  +  2  =  0. 

12.  V^+1  +  V:rTl6  =  V^+25. 

13.  V2^+l  -  V^Ti  =  i  V^^^. 

14.  V^  +  3  +  V^  +  8  =  bVx. 

15.    V3+^+V^=    ,J 16.    V^=I+6=      -^^ 


V3+^-  V^=I' 

1  ^    __        1 

^^*    V^H^      Va;-1      V:r2-l' 


,o     V:^  +  2a— Vo;  — 2a_  a; 

18.      ,  ,  : ~      • 

V:r-2a  +  V:r  +  2a      2a 
3a7+V4^^=^^      ^  V7.r^  +  4  +  2V3a;-l 


21.    V(:r  -ay  +  2ab  +  b^  =  x-a  +  b. 


22.    V(a;  +  a)2  +  2a^>  +  Z^2=-5-a-rr. 


RADICAL    EXPRESSIONS.  257 

24.  4:X^-Z (x^  +  1)  (:r^  ~2)=x^  (10  -  3^^). 

25.  (:rf-2)(^t_4)  =  J-(a;t-l)2-12. 

26.  ^^~4a:t  =  96.  28.   x^- +2o?x-^  =  ^a. 

27.  x^x-'--=2.Si.  29.   81-^  +  -^  =  52a;. 

290.    Equations  may  be  solved  with,  respect  to  an  expres- 
sion  in  the  same  manner  as  with  respect  to  a  letter. 

(1)    Solve  (x""  -xy-8(x'-  x)  +  12  =  0. 

Consider  {a^  —  x)  as  the  unknown  quantity. 
Then  {f  -  xf -?>{j^  -  x)  =  -  12. 

Complete  the  square,  (a;*  —  a;)'  —  (  )  +  16  =  4. 
Extract  the  root,  (x*  —  a;)  —  4  =  db  2. 

a^  -  a;  =  6  or  2. 
Complete  the  square,  4  a;*  —  ( )  +  1  =  25  or  9. 

Extract  the  root,  2  a;  —  1  =  ±  5  or  .-t  3. 

2a;=6, -4,  4, -2. 
.-.  a;=3, -2,  2, -1. 


C2)    Solve  5a;-7:r2-8V7ar»-5a;+l  =  8. 

Change  the  signs  and  annex  +  1  to  both  sides. 

7a^  _  5a;  +  1  +  8  V7ar»  -  5a;  +  1  =  -  7. 
Solve  -with  respect  to  VTx'  —  5  a;  +  1. 

(7a?-5a;  +  1)  +  8  (7ar'-5a;  +  1)^  +  16  =  9. 
(7ar'-5a;  +  l)^  +  4  =  ±3. 

(7a;2_5a;  +  l)5  =  _lor-7. 
Square,  7a;*  —  5  a;  +  1  =  1  or  49. 

Transpose,  73:*  —  5  a;  =  0  or  48. 

From  7a:^  — 5a;  =  0,         a;  =  Oor|^; 
From  7a;* -5  a;  =  48,       a;  =  3or-2f 

Note.  In  verifying  the  values  of  x  in  the  original  equation,  it  is 
seen  that  the  value  of  V7a;*  —  5  a;  +  1  is  negative.  Thus,  by  putting 
0  for  X  the  equation  becomes  0  —  8vT=  8 ;  and  by  taking  —  1  for  Vl 
we  have  (- 8)(- 1)  =  8  ;  that  is,  8  =.  8. 


258  ALGEBRA. 


(3)    ^olve  or" +  x +1  +  ^  +  ^=1. 
X     or 


Arrange  as  follows 

^(-.^)*("a- 

=  0. 

By  adding  2  to  fa.-* 

'^> 

.ere  is  obtained 

'*'-*H"'l 

H-*i)M"l)- 

Multiply  by  4  and 

complete  the  square, 

/ 

tf^  +  -Y+()  +  l=9. 

Extract  the  root, 

2(..l).i  =  .: 

3. 

2(..1)  =  . 

or -4. 

.  +  1  =  1 

or -2. 

X 

Multiply  by  a:,     a;*  —  a;  =»  —  1,  and        rc^  +  2  a;  =>  —  1. 

.•.4a!«-()  +  l  =  -3,  .•.ar«  +  2a:  +  l  =  0. 

2a;-l  =  ±V^,  a +  1  =  0. 

:.x=l{l±V-2>).  :.x==-l. 

291.  An  equation  like  that  of  (3)  which  will  remain  un- 
altered when  -  is  substituted  for  x,  is  called  a  reciprocal 

X 

equation. 

It  will  be  found  that  every  reciprocal  equation  of  odd  degree  will 
be  divisible  by  a;  —  1  or  a;  +  1  according  as  the  last  term  is  negative 
or  positive  ;  and  every  reciprocal  equation  of  even  degree  with  its  last 
term  negative  will  be  divisible  by  a;^  —  1.  In  every  case  the  equation 
resulting  from  the  division  will  be  reciprocal. 


RADICAL    EXPRESSIONS.  259 

(4)    Qolvea^  +  2x*-S3^-Sx^  +  2x  +  l  =  0. 

This  is  a  reciprocal  equation,  for,  if  x~^  be  put  for  x,  the  equation 
becomes  x~^  +  2x-*  —  3x-^  —  3x-^  +  2x-^  +  1  =  0,  which  multiplied 
by  x"  gives  l+2a;  —  3ic*  —  3a:^  +  2a;*  +  a^  =  0,  the  same  as  the  original 
equation. 

The  equation  may  be  written  (.r*  +  1)  +  2ic  (x^  +  1)  —  3  a;^  (x  +  1)  =  0, 
which  is  obviously  divisible  by  a;  +  1.  The  result  from  dividing  by 
a;+l  is  a;*  +  a;^  — 4ar^  +  a;  +  l  =  0,  or(.'C*  +  l)  +  a;(ar'+l)  =  4a'2.  By  add- 
ing 2^-2  to  {^  +  1)  it  becomes  (a;^  +  2ar^  +  1)  =  (a;^  +  I)'. 

Then  (ar^  +  l)^  +  a;  (x*  +  1)  =  6a;«. 

Multiply  by  4  and  complete  the  square, 

4(a^  +  l)2  +  (  )  +  ar»  =  25a^. 

Extract  the  root,  2 (a^  +  1)  +  a;  =  ±  5a;. 

Hence,  2  a;*^  +  2  =  4  a;  or  —  6  a?. 

By  simplifying,  ar*  — 2a;  =  —1;  and  a;'  +  3a;  =  — 1, 

whence,  a;  =  1  and  1 ;     whence,  a;  =  J  (—  3  ±  VS). 

Therefore,  including  the  root  —  1  obtained  from  the  factor 
a;  +  1,  theyiue  roots  are  —  1,  1,  1,  ^(— 3±  Vo). 

By  this  process  a  reciprocal  cubic  equation  may  be  reduced  to  a 
quadratic,  and  one  of  the  fifth  or  sixth  degree  to  a  biquadratic,  the 
solution  of  which  may  be  easily  effected. 

Solve  •  Exercise  CVIII. 

1.  a,'2-3.r-6Vx2-3:r-3  +  2  =  0. 

2.  ^  +  3a:-?  +  ^2  =  A- 

3.  {2x'~-?>xf-2{2x'-2>x)^lb. 

4.  {ax-bf-\-^a{ax-~h)  =  ^. 

5.  ?>{2x''~x)-{2o(^-x)^  =  2. 

6.  15.T-3:r2  +  4(:r2 -5:r  +  5)^=-16. 

7.    x^^x-^-\-x+x-^  =  4:.        9.    a^-{-x  +  ^{x'  +  xy^  =  l. 
a   a;2+ V:^'=^  =  19.  10.    (a;+l)^  +  (^- 1)^  =  5. 


260  ALGEBRA. 


11.   :p-l  =  2  +  25:-i  12.    V3^T5- V3a;-5  =  4. 

13.  (x'-\-l)-x(x'+l')  =  -~.2x'. 

14.  2x'-2^2:i^-bx  =  5(x+S). 

15.  x  +  2-4:xVx  +  2  =  12x^. 

16.  -^2x-\~a  +  -V2x-a  =  b. 

17.  ■V9x'  +  2lx-^l--V9x'  +  Qx+l=:Sx. 

18.  :t;l  — 4^t  +  ar-t  +  4^-t  =  --J. 

19.  (2a:  +  3y)2-2(2a;  +  32/)  =  8| 
2^2-2/^=^21  J 

20.   ^4- 2/+ V^  +  y  =  a|  22.    a^ -}- f  +  x -{:_7/ =  iS  ) 

x  —  y-{-  ^x  —  y=^h}  rr?/=12  J 

a?  —  xy -\- 'i/^  —  ?>      J  a;  +  V^y  +  y  = 

24.    (:r-yf-3(a;-y)  =  10| 


25.    ■\/x  —  Vy  =  a:^  {x^  +  y^)  ) 
(:.  +  yf=2(^-y)2  J 

» (;fj)'+C-s^)'=i 

rry  —  (a;  +  y)  =  54  ' 

27.  a7  +  y+V^  =  28| 
a.^  +  y2-|-a;y  =  336J 

28.  --3aa;=  V4PT9^  +  ^'. 
a  4 

29.  (a;  +  l+^'')(^-l+^"')  =  5i. 

30.  2  (:c^  -  1)-^  -  2  (a:^  -  4)-^  =  3  {x^^  -  2)-\ 


CHAPTER  XIX. 
Logarithms. 

292.  In  the  common  system  of  notation  the  expression 
of  numbers  is  founded  on  their  relation  to  ten. 

Thus,  3854  indicates  that  this  number  contains  10^  three  times,  10* 
eight  times,  10  five  times,  and  four  units. 

293.  In  this  system  a  number  is  represented  by  a  series 
of  different  powers  of  10,  the  exponent  of  each  power  being 
integral.  But,  by  employing /rac^^onaZ  exponents,  any  num- 
ber may  be  represented  (approximately)  as  a  single  power 
of  10. 

294.  When  numbers  are  referred  in  this  way  to  10,  the 
exponents  of  the  powers  corresponding  to  them  are  called 
their  logarithms  to  the  base  10. 

For  brevity  the  word  "logarithm"  is  written  log. 

From  §  255  it  appears  that : 

10"  =     1,  10->(=tV)     =1. 

10'=  10,  io-^(=Tk)  =-01, 

W=100,  10-«(=„i^)  =  .001, 

and  so  on.     Hence, 

log      1  =  0,  log.l      =-1, 

log    10  =  1,  log  .01    =-2, 

log  100  =  2,  log  .001  =-3, 
and  so  on. 


262  ALGEBRA. 


It  is  evident  that  the  logarithms  of  all  numbers  between 

1  and      10  will  be       0  +  a  fraction, 

10  and    100  will  be       1  +  a  fraction, 

100  and  1000  will  be       2  +  a  fraction, 

1  and   .1      will  be  —  1  +  a  fraction, 

.1    and   .01    will  be  —  2  +  a  fraction, 

.01  and   .001  will  be  —  3  +  a  fraction. 

295.  The  fractional  part  of  a  logarithm  cannot  be  ex- 
pressed exactly  either  by  common  or  by  decimal  fractions  ; 
but  decimals  may  be  obtained  for  these  fractional  parts, 
true  to  as  many  places  as  may  be  desired. 

If,  for  instance,  the  logarithm  of  2  be  required  ;  log  2  may  be  sup- 
posed to  be  ^. 

Then  10^  =  2 ;  or,  by  raising  both  sides  to  the  third  power,  10  =  8, 
a  result  which  shows  that  \  is  too  large.  ^ 

Suppose,  then,  log  2  =  -^q.  Then  10^"  =  2,  or  by  raising  both  sides 
to  the  tenth  power,  10^  =  2^°.  That  is,  1000  =  1024,  a  result  which 
shows  that  -^^  is  too  small. 

Since  ^  is  too  large  and  j^^  too  small,  log  2  lies  between  J  and  y^^  ; 
that  is,  between  .33333  and  .30000. 

In  supposing  log  2  to  be  J,  the  error  of  the  result  is  if  ^  =  ^g-  =  .2. 
In  supposing  log  2  to  be  -j^,  the  error  of  the  result  is  -^^xV^yr^^  ^ 
Y^2_4j.  =  _  .024 ;  log  2,  therefore,  is  nearer  to  -^^^  than  to  ^. 

The  difference  between  the  errors  is  .2  —  (—  .024)  =  .224,  and  the 
difference  between  the  supposed  logarithms  is  .33333  —  .3  =  .03333. 

The  last  error,  therefore,  in  the  supposed  logarithm  may  be  consid- 
ered to  be  approximately  ^W  of  .03333  =  .0035  nearly,  and  this  added 
to  .3000  gives  .3035,  a  result  a  little  too  large. 

By  shorter  methods  of  higher  mathematics,  the  logarithm  of  2  is 
known  to  be  0.3010300,  true  to  the  seventh  place. 

296.  The  logarithm  of  a  number  consists  of  two  parts,  an 
integral  part  and  a  fractional  part. 

Thus,  log  2  =  0.30103,  in  which  the  integral  part  is  0,  and  the  frac- 
tional part  is  .30103  ;  log  20  =  1.30103,  in  which  the  integral  part  is  1, 
and  the  fractional  part  is  .30103. 


LOGARITHMS.  263 


297.  The  integral  part  of  a  logarithm  is  called  the  char- 
acteristic ;  and  the  fractional  part  is  called  the  mantissa. 

298.  The  mantissa  is  always  made  positive.  Hence,  in 
the  case  of  numbers  less  than  1  whose  logarithms  are  nega- 
tive, the  logarithm  is  made  to  consist  of  a  negative  charac- 
teristic and  a  positive  mantissa. 

299.  When  a  logarithm  consists  of  a  negative  character- 
istic and  a  positive  mantissa,  it  is  usual  to  write  the  minus 
sign  over  the  characteristic,  or  else  to  add  10  to  the  charac- 
teristic and  to  indicate  the  subtraction  of  10  from  the 
resulting  logarithm. 

Thus,  log  .2  =  T.30103,  and  this  may  be  written  9.30103  - 10. 

300.  The  character istie  of  a  logarithm  of  an  integral  num- 
ber, or  of  a  mixed  number,  is  one  less  than  the  number  of 
integral  digits. 

Thus,  from  ^  294,  log  1  =  0,  log  10  =  1,  log  100  =  2.-  Hence,  the 
logarithms  of  all  numbers  from  1  to  10  (that  is,  of  all  numbers  con- 
sisting of  one  integral  digit),  will  have  0  for  characteristic ;  and  the 
logarithms  of  all  numbers  from  10  to  100  (that  is,  of  all  numbers  con- 
sisting of  two  integral  digits),  will  have  1  for  characteristic  ;•  and  so 
on,  the  characteristic  increasing  by  1  for  each  increase  in  the  number 
of  digits,  and  therefore  always  being  1  less  than  that  number. 

301.  Tlte  characteristic  of  a  logarithm  of  a  decimal  frac- 
tion is  negative,  and  is  equal  to  the  number  of  the  place  occu- 
pied by  the  first  significant  figure  of  the  depimal. 

Thus,  from  \  294,  log  .1  =  - 1,  log  .01  =  -2,  log  .001  =  -3.  Hence, 
the  logarithms  of  all  numbers  from  ,1  to  1  will  have  —  1  for  a  charac- 
teristic (the  mantissa  being  "plui) ;  the  logarithms  of  all  numbers  from 
.01  to  .1  will  have  —  2  for  a  characteristic ;  the  logarithms  of  all  num- 
bers from  .01  to  .001  will  have  —  3  for  a  characteristic ;  and  so  on, 
the  characteristic  always  being  negative  and  equal  to  the  number  of  thi 
place  occupied  hy  the  first  significant  figure  of  the  decimal. 


264  ALGEBRA. 


302.  The  mantissa  of  a  logarithm  of  any  integral  num- 
ber or  decimal  fraction  depends  only  upon  the  digits  of  the 
number,  and  is  unchanged  so  long  as  the  sequence  of  the 
digits  remains  the  same. 

For,  changing  the  position  of  the  decimal  point  in  a  number  is 
equivalent  to  multiplying  or  dividing  the  number  by  a  power  of  10. 
Its  logarithm,  therefore,  will  be  increased  or  diminished  by  the  expo- 
nent of  that  power  of  10 ;  and,  since  this  exponent  is  integral,  the 
mantissa  of  the  logarithm  will  be  unaffected. 

Thus,  if  27196  =  lO^-^*^, 

then  2719.6  =  lO^-^^s^ 

27.196  =  lO^'"^, 

2.7196  =  W'*^, 

.27196  =   109-4345-W 

.0027196  =  10' ««-»«. 

303.  The  advantage  of  using  the  number  10  ap,  the  ba.«e 
of  a  system  of  logarithms  consists  in  the  fact  taat  the  man- 
tissa depends  only  on  the  sequence  of  digits,  and  the  charac- 
teristic on  the  position  of  the  decimal  point. 

304.  As  logarithms  are  simply  exponents  (§  294),  therefore, 
The  logarithm  of  a  product  is  the  sum  of  the  logarithms 

of  the  factors. 

Thus,  log  20  =  log  (2  X  10)  =  log  2  +  log  10 

=  0.3010  +  1.0000  =  1.3010; 
log  2000  =  log  (2  X  1000)  =  log  2  +  log  1000, 

=  0.3010  +  3.0000=3.3010; 
log  .2  =  log  (2  X  .1)  =  log  2  +  log  .1, 

=  0.3010  +  9.0000  -  10  -  9.3010  -  10 ; 
log  .02  =  log  (2  X  .01)  =  log  2  +  log  .01, 

=  0.3010  +  8.0000  -  10  =  8.3010  -  10. 

Exercise  CIX. 

Given:  log  2  =  0.3010;  log  3  =  0.4771;  log  5  =  0.6990; 
log  7  =  0.8451. 

Find  the  logarithms  of  the  following  numbers  by  resolv- 


LOGABITHMS. 

265 

ing  the  numbers  into  factors, 

and  taking 

the  sum  of  the 

logarithms 

of  the  factors : 

1.   log  6. 

9.   log  25. 

17. 

log  .021. 

25. 

log  2.1. 

2.    log  15. 

10.    log  30. 

18. 

log  .35. 

26. 

log  16. 

3.    log  21. 

11.   log  42. 

19. 

log  .0035. 

27. 

log  .056. 

4.   log  14. 

12.   log  420. 

20. 

log  .004. 

28. 

log  .63. 

5.   log  35. 

13.    log  12. 

21. 

log  .05. 

29. 

log  1.75. 

6.    log  9. 

14.   log  60. 

22. 

log  12.5. 

30. 

log  105. 

7.   log  8. 

15.   log  75. 

23. 

log  1.25. 

31. 

log  .0105. 

8.   log  49. 

16.    log  7.5. 

24. 

log  37.5. 

32. 

log  1.05. 

305.  As  logarithms  are  simply  exponents  (§  294),  there- 
fore, 

TJie  logarithm  of  a  power  of  a  number  is  equal  to  the 
logarithm  of  the  number  Tnultiplied  by  the  exponent  of  the 
power. 

Thus,  log  5'  =    7  X  log  5=    7x0.6990  =  4.8930. 

log  3"  =  11  X  log  3  =  11  X  0.4771  =  5.2481. 

306.  As  logarithms  are  simply  exponents  (§  294),  there- 
fore, when  roots  are  expressed  by  fractional  indices, 

The  logarithm  of  a  root  of  a  number  is  equal  to  the  loga- 
rithm of  the  number  multiplied  by  the  index  of  the  root. 

Thus,  log  2^      =  1  of  log  2  =  1  X  0.3010  =  0.0753. 

log  .002^  =  J  of  (7.3010  -  10). 

The  expression  \  of  (7.3010  -  10) 

may  be  put  in  the  form  of  \  of  (27.3010  -  30)  which  =  9.1003  -  10 ; 
for,  since  20  -  20  =  0,  the  addition  of  20  to  the  7,  and  of  -  20  to  the 
—  10,  produces  no  change  in  the  value,  of  the  logarithm. 

307.  In  simplifying  the  logarithm  of  a  root  the  equal  pos- 
itive and  negative  numbers  to  be  added  to  the  logarithm  must 
be  such  that  the  resulting  negative  number,  when  divided  by 
the  index  of  the  root,  shall  give  a  quotient  of  —  10. 


266  ALGEBRA. 


Exercise  CX. 

Given:  log  2  =  0.3010 ;  log  3  =  0.4771 ;  log  5  =  0.6990; 
los  7  =  0.8451. 


Find  logarithms  of  the  following : 


1.   2\ 

6.    5^ 

11. 

5^. 

16.    7f. 

21.    5i 

2.   b\ 

7.   2i. 

12. 

7tV. 

17.    5t. 

22.    2V. 

3.    1\ 

8.    5i 

13. 

2l. 

18.    3t't. 

23.    5J. 

4.  38. 

9.    3i 

14. 

5*. 

19.       72. 

24.    7-V, 

5.    7^ 

10.   7-^ 

15. 

3f 

20.    3i 

25.    21^, 

308.  Since  logarithms  are  simply  exponents  (§  294),  there- 
fore, 

The  logarithm  of  a  quotient  is  the  logarithm  of  the  divi- 
dend minus  the  logarithm  of  the  divisor. 

Thus,      log  I  =  log  3  -  log  2  =  0.4771  -  0.3010  =     0.1761. 
log  I  =  log  2  -  log  3  -  0.3010  -  0.4771  =  -  0.1761. 

To  avoid  the  negative  logarithm  —  0.1761,  we  subtract  the  entire 
logarithm  0.1761  from  10,  and  then  indicate  the  subtraction  of  10 
from  the  result. 

Thus,  -  0.1761  =  9.8239  -  10. 

Hence,  log  f  =  9.8239  -  10. 

309.  The  remainder  obtained  by  subtracting  the  loga- 
rithm of  a  number  from  10  is  called  the  cologarithm  of  the 
number,  or  arithmetical  complement  of  the  logarithm  of  the 
number. 

Cologarithm  is  usually  denoted  by  colog,  and  is  most 
easily  found  by  beginning  with  the  characteristic  of  the  loga- 
rithm and  subtracting  each  figure  from  9  down  to  the  last 
significant  figure,  and  subtracting  that  figure  from  10. 

Thus,  log  7  =  0.8451 ;  and  colog  7  =  9.1549.  Colog  7  is  readily 
found  by  subtracting,  mentally,  0  from  9,  8  from  9,  4  from  9,  5  from 
9,  1  from  10,  and  writing  the  resulting  figure  at  each  step. 


LOGARITHMS.  267 


310.  Since  colog  7  =  9.1549, 

and       log  f  =  log  1-log  7  =  0-0.8451  =  9.1549-10, 
it  is  evident  that, 

IflObe  subtracted  from  the  cologarithm  of  a  number,  the 
•result  is  the  logarithm  of  the  reciprocal  of  that  number. 

311.  Since  log  |  =  log  7  —  log  5, 

=  0.8451  -  0.G990  =  0.1461. 
and    log  7  +  colog  5  -  10  =  0.8451  +  9.3010  -  10, 

=  0.1461, 
it  is  evident  that, 

The  addition  of  a  cologarithm  —  10  is  equivalent  to  the 
subtraction  of  a  logarithm. 

The  steps  that  lead  to  this  result  are: 

therefore,  log  |  =  log  (7  X  i)  =  log  7  +  log  i.  g  304. 

But  log  \  =  colog  5-10.  I  309. 

Hence,  log  -^  =  log  7  +  colog  5  —  10. 

Therefore, 

312.  The  logarithm  of  a  quotient  may  be  found  by  add- 
ing together  the  logarithm  of  the  dividend  and  the  cologa- 
rithm of  the  divisor,  and  subtracting  10  from  the  result. 

In  finding  a  cologarithm  when  the  characteristic  of  the  logarithm 
is  a  negative  number,  it  must  be  observed  that  the  subtraction  of  a 
negative  number  is  equivalent  to  the  addition  of  an  equal  positive 
number. 

Thus,  log  —  =  log  5  +  colog  .002  -  10, 

°  .002 

=  0.6990  +  12.6990  -  10, 

=  3.3980. 

Here  log  .002  =  3.3010,  and  in  subtracting  -  3  from  9  the  result  h 

the  same  as  adding  +  3  to  9. 
9 
Again,  log  -^  =  log  2  +  colog  .07  —  10, 

=  0.3010  +  11.1549  -  10, 
=  1.4559. 


268  ALGEBRA. 


Also, 

log  --^  =  8.8451  -  10  +  9.0970  -  10, 
=  17.9421  -  20, 

=  7.9421-10. 

Here, 

log  23  =  3  log  2  =  3  X  0.3010  =  0.9030. 

Hence, 

colog  2^  =  10  -  0.9030  =  9.0970. 

- 

Exercise  CXI. 

Given:  log  2-0.3010;  log  3  =  0.4771;  log  5  =  0.6990; 
log  7  =  0.8451. 

Find  logarithms  for  the  following  quotients : 

1.  I            7.   ^.  13.   •-^.  19.  ^.  25.   -^. 

5  3  3  .003  33 

2.  2             8.  ^  14.  -^  20    -^  26.  -^ 

7                   2'  2  '  .02"  .02*' 

3.  5.             9.  I  15.  ^.  21.  ^.  27.   -^, 

6  3  5  .007  .022 

4.  -.           10.  -.  16.   — .  22.  — .  28.  -^. 

7  2  .07  .07  .0033 

6.  I           11.  ^.  ^   17.  _^.  23.  :^.  29.  ^ 

7                  2  .007  7  73 

6.  X           12.  X  18.  :^^X  24.   :^007  3^^  JZl 

5                   .5  7  .2  .005* 


313.  A  table  of  four-place  logarithms  is  here  given,  which 
contains  logarithms  of  all  numbers  under  1000,  the  decimal 
point  and  characteristic  being  omitted.  The  logarithms  of 
single  digits  1,  8,  etc.,  will  be  found  at  10,  80,  etc. 

Tables  containing  logarithms  of  more  places  can  be  pro- 
cured, but  this  table  will  serve  for  many  practical  uses,  and 
will  enable  the  student  to  use  tables  of  six-place,  seven- 
place,  and  ten-place  logarithms,  in  work  that  requires 
greater  accuracy. 


LOGARITHMS.  269 


314.  In  working  with,  a  four-place  table,  the  numbers 
corresponding  to  the  logarithms,  that  is,  the  antilogarithiyis, 
as  the  J  are  called,  may  be  carried  to  four  significant  digits. 

.To  Find  the  Logarithm  of  a  Number  in  this  Table. 

315.  Suppose  it  is  required  to  find  the  logarithm  of  65.7. 
In  the  column  headed  "  N"  look  for  the  first  two  significant 
figures,  and  at  the  top  of  the  table  for  the  third  significant 
figure.  In  the  line  with  65,  and  in  the  column  headed  7, 
is  seen  8176.  To  this  number  prefix  the  characteristic  and 
insert  the  decimal  point.     Thus, 

log  65.7  =  1.8176. 

Suppose  it  is  required  to  find  the  logarithm  of  20347. 
In  the  line  with  20,  and  in  the  column  headed  3,  is  seen 
3075  ;  also  in  the  line  with  20,  and  in  the  4  column,  is  seen 
3096,  and  the  difierence  between  these  two  is  21.  The  dif- 
ference between  20300  and  20400  is  100,  and  the  difierence 
between  20300  and  20347  is  47.  Hence,  -j^  of  21  =  10, 
nearly,  must  be  added  to  3075.  That  is, 
log  20347  =  4.3085. 

Suppose  it  is  required  to  find  the  logarithm  of  .0005076. 
In  the  line  with  50,  and  in  the  7  column,  is  seen  7050 ;  in 
the  8  column,  7059 :  the  difierence  is  9.  The  difierence  be- 
tween 5070  and  5080  is  10,  and  the  diff'erence  between  5070 
and  5076  is  6.  Hence,  ^^  of  9  =  5  must  be  added  to  7050. 
That  IS,  ^^g  .0005076  =  6.7055  -  10. 

To  Find  a  Number  when  its  Logarithm  is  Given. 

316.  Suppose  it  is  required  to  find  the  number  of  which 
the  logarithm  is  1.9736. 

Look  for  9736  in  the  table.  In  the  column  headed  "N," 
and  in  the  line  with  9736,  is  seen  94,  and  at  the  head  of 


270 


ALGEBRA. 


N 

0 

1   2 

3 

4 

5   6 

7 

8 

9 

10 

0000 

0043 

0036 

0128 

017010212 

0253 

0294 

0334 

0374 

11 

0414 

0453 

0192 

0531 

0569 

0007 

0645 

0682 

0719 

0755 

12 

0792 

0823 

0864 

0899 

0934 

0969 

1004 

1033 

1072 

1106 

13 

1139 

1173 

1203 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

15 

1701 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

10 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2705 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3902 

2o 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4450 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

29 
30 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5070 

37 

56S2 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

38 

5798 

5809 

5321 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977  5988 

5999 

6010 

40 

6021 

6031 

6042 

6053 

"6064 

6075 

6085 

6096 

6107 

6117 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

43 

6335 

6345 

6355 

6365 

6375 

6385 

0395 

6405 

6415 

6425 

44 

6435 

6444 

6454 

6464 

6474 

0484 

0493 

6503 

6513 

6522 

45^ 

6532 

6542 

6551 

6531 

6571 

0580 

0590 

6599 

6609 

¥018 

46 

6028 

6637 

6646 

6656 

6665 

0075 

6684 

6693 

6702 

6712 

47 

6721 

6730 

6739 

0749 

6758 

6767 

6776 

6785 

6794 

6803 

48 

6812 

6821 

6830 

6339 

6848 

6857 

C866 

6875 

6884 

6S93 

49 

6902 

6911 

6920 

6923 

6937 

0946 

6955 

6904 

6972 

0981 

50 

6990 

6993 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7007 

51 

7070 

7084 

7093 

7101 

7110 

7118 

7126 

7135  7143 

7152 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

LOGARITHMS. 


271 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8   9 

55 

56 
57 
58 
59 

7404 
7482 
7559 
7634 
7709 

7412 
7490 
7566 
7642 
7716 

7419 
7497 
7574 
7649 
7723 

7427 
7505 
7582 
7657 
7731 

7435 
7513 
7589 
7664 
7738 

7443 
7520 
7597 
7672 
7745 

7451 
752S 
7604 
7679 
7752 

7459 
7536 
7612 
7086 
7760 

7466 
7543 
7619 
7694 
7767 

7474 
7551 
7627 
7701 
7774 

60 

61 
62 
63 
64 

7782 
7853 
7924 
7993 
8062 

7789 
7860 
7931 
8000 
8069 

7796 
7868 
7938 
8007 
8075 

7803 
7875 
7945 
8014 
8082 

7810 
7882 
7952 
8021 
8089 

7818 
7889 
7959 
8028 
8096 

7825 
7896 
7966 
8035 
8102 

7832 
7903 
7973 
8041 
8109 

7839 
7910 
7980 
8048 
8116 

7846 
7917 
7987 
8055 
8122 

65 

66 
67 
68 
69 

8129 
8195 
8261 
8325 
8388 

8136 
8202 
8267 
8331 
8395 

8142 
8209 
8274 
8338 
8401 

8149 
8215 
8280 
8344 
8407 

8156 
8222 
8287 
8351 
8414 

8162 
8228 
8293 
8357 
8420 

8169 
8235 
8299 
8363 
8426 

8176 
8241 
8306 
8370 
8432 

8182 
8248 
8312 
8376 
8439 

8189 
8254 
8319 
8382 
8445 

70 

71 
72 
73 
74 

8451 
8513 
8573 
8633 
8692 

8457 
8519 
8579 
863« 
8698 

8463 
8525 
8585 
8645 
8704 

8470 
8531 
8591 
8651 
8710 

8476 
8537 
8597 
8657 
8716 

8482 
8543 
8603 
8663 

8722 

8488 
8549 
8609 
8669 
8727 

8494 
8555 
8615 
8675 
8733 

8500 
8561 
8621 
8681 
8739 

8506 
8567 
8627 
8686 
8745 

75 

76 
77 
78 
79 

8751 
8808 
8865 
8921 
8976 

8756 
8814 
8871 
8927 
8982 

8762 
8820 
8876 
8932 
8987 

8768 
8825 
8882 
8938 
8993 

8774 
8831 
8887 
8943 
8998 

8779 
8837 
8893 
8949 
9004 

8785 
8842 
8899 
8954 
9009 

8791 
8848 
8904 
8960 
9015 

8797 
8854 
8910 
8965 
9020 

8802 
8859 
8915 
8971 
9025 

80 
81 
82 
83 
84 

9031 
9085 
9138 
9191 
9243 

9036 
9090 
9143 
9196 
9248 

9042 
9096 
9149 
9201 
9253 

9047 
9101 
9154 
9206 
9258 

9053 
9106 
9159 
9212 
9263 

9058 
9112 
9165 
9217 
9269 

9063 
9117 
9170 
9222 
9274 

9069 
9122 
9175 
9227 
9279 

9074 
9128 
9180 
9232 

9284 

9079 
9133 
9186 
9238 
9289 

85 

86 
87 
88 
89 

9294 
9345 
9395 
9445 
9494 

9299 
9350 
9100 
9450 
9499 

9304 
9355 
9405 
9455 
9504 

9309 
9360 
•9410 
9460 
9509 

9315 
9365 
9415 
9465 
9513 

9320 
9370 
9420 
9469 
9518 

9325 
9375 
9425 
9474 
9523 

9330 
9380 
9430 
9479 

9528 

9335 
9385 
9435 
9484 
9533 

9340 
9390 
9440 
9489 
9538 

90 

91 
92 
93 
94 

9542 
9590 
9638 
9685 
9731 

9547 
9595 
9643 
9689 
9736 

9552 
9600 
9647 
9694 
9741 

9557 
9605 
9652 
9699 
9745 

9562 
9609 
9657 
9703 
9750 

9566 
9614 
9661 
9708 
9754 

9571 
9619 
9666 
9713 
9759 

9576 
9624 
9671 
9717 
9763 

9581 
9628 
9675 
9722 
9768 

9586 
9633 
9680 
9727 
9773 

95 

98 
97 
98 
99 

9777 
9823 
9868 
9912 
9956 

9782 
9827 
9872 
9917 
9961 

9786 
9832 
9877 
9921 
9965 

9791 
9836 
9881 
9926 
9969 

9795 
9841 
9886 
9930 
9974 

9800 
9845 
9890 
9934 
9978 

9805 
9850 
9894 
9939 
9983 

9809 
9854 
9899 
9943 
9987 

9814 
9859 
9903 
9948 
9991 

9818 
9863 
9908 
9952 
9996 

272  ALffEBEA. 

the  column  in  which  9736  stands  is  seen  1.  Therefore, 
write  941,  and  insert  the  decimal  point  as  the  characteristic 
directs.     That  is,  the  number  required  is  94.1. 

Suppose  it  is  required  to  find  the  number  of  which  the 
logarithm  is  3.7936. 

Look  for  7936  in  the  table.  It  cannot  be  found,  but  the 
two  adjacent  mantissas  between  which  it  lies  are  seen  to  be 
7931  and  7938 ;  their  difference  is  7,  and  the  difference  be- 
tween 7931  and  7936  is  5.  Therefore,  f  of  the  difference 
between  the  numbers  corresponding  to  the  mantissas,  7931 
and  7938,  must  be  added  to  the  number  corresponding  to 
the  mantissa  7931. 

The  number  corresponding  to  the  mantissa  7938  is  6220. 

The  number  corresponding  to  the  mantissa  7931  is  6210. 

The  difference  between  these  numbers  is  10, 
and  6210-f-f  of  10  =  6217. 

Therefore,  the  number  required  is  6217. 

Suppose  it  is  required  to  find  the  number  of  which  the 
logarithm  is  7.3882  —  10. 

Look  for  3882  in  the  table.  It  cannot  be  found,  but  the 
two  adjacent  mantissas  between  which  it  lies  are  seen  to  be 
3874  and  3892;  their  difference  is  18,  and  the  difference 
between  3874  and  3882  is  8.  Therefore,  ^  of  the  differ- 
ence between  the  numbers  corresponding  to  the  mantissas, 
3874  and  3892,  must  be  added  to  the  number  corresponding 
to  the  mantissa  3874. 

The  number  corresponding  to  the  mantissa  3892  is  2450. 
The  number  corresponding  to  the  mantissa  3874  is  2440. 
The  difference  between  these  numbers  is  10, 
and  2440  + 1  of  10  =  2444. 

Therefore,  the  number  required  is  .002444. 


logarithms.  '  273 

Exercise  CXII. 
Find  logaritlims  of  the  following  numbers  : 

1.  60.  6.   3780.  11.    70633.  16.  877.08. 

2.  101.  7.    54327.  12.    12028.  17.  73.896. 

3.  999.  a    90801.  13.    0.00987.  18.  7.0699. 

4.  9901.  9.   10001.  14.   0.87701.  19.- 0.0897. 

5.  5406.  10.    10010.  15.    1.0001.  20.  99.778. 

Find  antilogarithms  to  the  following  logarithms : 

21.  4.2488.  25.   4.7317.  29.   9.0410-10. 

22.  3.6330.  26.    1.9730.  30.    9.8420-10. 

23.  2.5310.  27.   9.8800-10.        31.    7.0216-10. 

24.  1.9484.  2a   0.2787.  32.   8.6580-10. 

Ex.    Find  the  product  of  908.4  X  .05392  X  2.117. 

log    908.4  =  2.9583 
log  .05392  =  8.7318  - 10 
log    2.117  =  0.3257 

2.0158  =  log  103.7.  Aiu. 

Find  by  logarithms  the  following  products : 

33.  948.76x0.043875.  35.  830.75x0.0003769. 

34.  3.4097x0.0087634.  36.   8.4395x0.98274. 

317.  When  any  of  the  factors  are  negative,  find  their  log- 
arithms without  regard  to  the  signs;  write  the  letter  n 
after  the  logarithm  that  corresponds  to  a  negative  number. 
If  the  number  of  logarithms  so  marked  be  odd,  the  product 
is  negative;  if  even,  the  product  ia positive. 


274 


ALGEBRA. 


Find  the  products  of : 
37.  7564  X  (-0.003764). 
S8.   3.7648 X (-0.083497). 

Ex.  Find  the  quotient  of 


39.  -5.840359 X  (-0.00178). 

40.  -8945.07x73.846. 

8.3709  X  834.637 


7308.946 
log  8.3709  =  0.9227 

log        834.637  =  2.9215 
colog  7308.946 


Find  the  quotients  of 
,,     70654 


41. 

54013' 

42. 

58706 
93078" 

43. 

8.32165 

0.07891" 

44. 

65039 

90761" 

AK. 

7.652 

0.06875 


46. 


47. 


48. 


49. 


50. 


6.1362-10 

9.9804 -10  =  log  .9558.  Ans. 


0.07054 


83.947  X  0.8395 

7564  X  0.07643 
8093  X  0.09817" 

89  X  753  X  0.0097 
36709  X  0.08497  " 

413  X  8.17  X  3182 
915  X  728  X  2.315' 

212  x(- 6.12)  x(- 2008) 
365  X  (- 531)  X  2.576    ' 


Ex.  Find  th 

e  cube  of  .0497. 

log  .0497  =  8.6964 - 
3 

■10 

6.0892  - 

•  10=  log. 

0001228.  Ans. 

Find  by  logarithms : 

51.    6.05^. 

55.    0.78765«. 

59. 

(10|)^ 

63.  (s^y-''- 

52.    1.05r. 

56.    0.69P. 

60. 

(IF- 

64.  (i^r- 

53.    1.1768^ 

57.  (i^r- 

61. 

(M¥f- 

65.    (8J)-'-^. 

54.    1.3178^'^. 

58.    (W'- 

62. 

(7_,_)0.3. 

\    66.    (5f|)»-«". 

LOGARITHMS.  275 


Ex.    Find  the  fourtli  root  of  0.00862. 

log  0.00862 

=    7.9355-10 

30.         -  30 

4)37.9355  -  40 

9.4839  -  10  =  log  .3047.  Ans. 

Find  by  logaritlims : 

67.  73. 

70.  .8379tV. 

73.  0.176431     76.  (^Vo^)'. 

68.   llK 

71.  906.804. 

74.  2.5637^'^.     77.  (9|-^)3. 

69.  7833. 

72.  8.1904i 

75.  m^.         78.  (llfi)- 

Find  by  logarithms  the  values  of: 

^g     ,10.0075433^  X  78.843  X  8172.4»  x  0.00052 
\  '  64285.^  X  154.27*  X  0.001  X  586.79* 


80. 


81. 


82. 


83. 


84. 


85. 


86. 


87. 


J        15.832«  X  5793.6^  X  0.78426 

\ 01)00327^  X  768.942  X  3015.3x0.007*' 

J7.1895  X  4764.2^  X  0.00326^ 
\0.00048953  X  457«  X  5764'42* 

3.1416  X  4771.21  x  2.7183* 


4 


30.103*  X  0.4343*  X  69.897* 

J  /0.032712  X  53.429  x  0.77542^ 
32.769  X  0.000371* 


\42 

4 


732.056^  X  0.0003572*  X  89793 
27983  X  3.4574  X  0.0026518^* 

7932x0.00657x0.80464 
0.03274  X  0.6428 

7.1206  X  Val3274  X  0.057389 


VO.43468  X  17.385  X  VO.0096372 

r  3.075526^  X  5771.2*  X  0.0036984vx  7.74  |* 
I  72258  X  327.93«  X  86.97^  J  ' 


276  ALGEBRA. 

318.  Since  any  positive  number  other  than  1  may  be 
taken  as  the  base  of  a  system  of  logarithms,  the  following 
general  proofs  to  the  base  a  should  be  noticed. 

I.  The  logarithm  of  the  product  of  two  or  more  numbers 
is  eqtial  to  the  sum  of  the  logarithms  of  the  numbers. 

For,  let  m  and  n  be  two  numbers,  and  x  and  y  their  logarithms. 
Then,  by  the  definition  of  a  logarithm,  m  =  a*  and  n  =  afl. 
Hence,  m  X  n  =  a*  X  a^  =  aj'+v. 

.'.  log  (m  X  n)  =  aj  +  y, 

=  log  m  +  log  n. 
In  like  manner,  the  proposition  may  be  extended  to  any  number 
of  factors. 

II.  The  logarithm  of  a  quotient  is  equal  to  the  logarithm 
of  the  dividend  minus  the  logarithm  of  the  divisor. 

For,  let  m  and  n  be  two  numbers,  and  x  and  y  their  logarithms. 
Then  m  =  a*  and  n  =  oM. 

Hence,  'm-i-n  =  a''  ^  av  =  a'-v. 

.'.  log  {m^n)  =  X  ~  y, 

=  log  m  —  log  n. 

From  this  it  follows  that  log  —  =  log  1  —  log  m. 
But,  since  log  1  =  0,  log  —  =  —  log  m. 

III.  The  logarithm  of  a  poiver  of  a  number  is  equal  to 
the  logarithm  of  the  number  m^ultiplied  by  the  exponent  of 
the  power. 

For,  let  X  be  the  logarithm  of  w. 

Then  m  =  a* 

and  mP  =  {a^y  =  aJ«. 

.•.  log  mP  =  px, 

=  p  log  m. 


LOGARITHMS.  277 


IV.    The  logarithm  of  the  root  of  a  number  is  equal  to  the 
logarithm  of  the  number  divided  by  the  index  of  the  root. 
For,  let  X  bo  the  logarithm  of  m. 

Then  m  =  a'', 

and  m^  =  (a*)    =  (V"- 

T  i        X        l02  TO 

&  r  r 

319.  An  exponential  equation,  that  is,  an  equation  in 
which  the  exponent  is  the  unknown  quantity,  is  easily 
solved  by  logarithms. 

For,  let  a*  =  TO. 

Then  log  a*  =-  log  m, 

.-.  X  log  a  =  log  w, 

.  ^  _  logm^ 

log  a* 

Ex.  Find  the  value  of  x  in  81="  =  10. 

8F  =  10, 
^_loglO 

.'.  log  X  =  log  log  10  +  colog  log  81, 
=  0  +  9.7193  -  10, 
.-.a;  =0.524. 

320.  Logarithms  of  numbers  to  any  base  a  may  be  con- 
verted into  logarithms  to  any  other  base  b  by  dividing  the 
computed  logarithms  by  the  logarithm  of  b  to  the  base  a. 

For,  let  log  rn  =  y  to  the  base  b, 

and  log   h  =  X  to  the  base  a. 

Then  to  =  Z>y,  and  b  =  a", 

.-.  m  =  (a'^y  =  a*y. 

.".  log  TO  (to  base  a)  ==  xy  =  log  b  (to  base  a)  X  log  m  (to  base  6), 

^  ,,    I         J.      lo;]r  TO  (to  base  a) 

.-.  log  TO  (to  base  6)  =  ,  ^  ,  ,\    , -^. 

°  log  6  (to  base  a) 

This  is  usually  written,  logs  w  —     ^* 


log«6 


CHAPTER  XX. 

Ratio,  Proportion,  and  Variation. 

321.  The  relative  "magnitude  of  two  numbers  is  called 
their  ratio,  and  is  expressed  by  the  fraction  which  the  first 
is  of  the  second. 

Thus,  the  ratio  of  6  to  3  is  indicated  by  the  fraction  |,  which  is 
sometimes  written  6 :  3. 

322.  The  first  term  of  a  ratio  is  called  the  antecedent, 
and  the  second  term  the  consequent.  When  the  antecedent 
is  equal  to  the  consequent,  the  ratio  is  called  a  o^atio  of 
equality ;  when  the  antecedent  is  greater  than  the  conse- 
quent, the  ratio  is  called  a  o^atio  of  greater  inequality ;  when 
less,  a  ratio  of  less  inequality. 

323.  When  the  antecedent  and  consequent  are  inter- 
changed, the  resulting  ratio  is  called  the  inverse  of  the  given 
ratio. 

Thus,  the  ratio  3 :  6  is  the  inverse  of  the  ratio  G :  3. 

324.  The  ratio  of  two  quantities  that  can  be  expressed  in 
integers  in  terms  of  a  common  unit  is  equal  to  the  ratio  of 
the  two  numbers  by  which  they  are  expressed. 

Thus,  the  ratio  of  $9  to  $11  is  equal  to  the  ratio  of  9  :  11 ;  and  the 
ratio  of  a  line  2f  inches  long  to  a  Hne  3  J  inches  long,  when  both  are 
expressed  in  terms  of  a  unit  jj  ^f  ^^  i'^^l^  long,  is  equal  to  the  ratio 
of  32  to  45. 

325.  Two  quantities  different  in  hind  can  have  no  ratio, 
for  then  one  cannot  be  a  fraction  of  the  other. 


RATIO.  279 


326.  Two  quantities  that  can  be  expressed  in  integers  in 
terms  of  a  common  unit  are  said  to  be  commensurable. 
The  common  unit  is  called  a  com^mon  Tneasure,  and  each 
quantity  is  called  a  multiple  of  this  common  measure. 

Thus,  a  common  measure  of  2n  feet  and  3f  feet  is  ^  of  a  foot,  which 
is  contained  15  times  in  2J  feet,  and  22  times  in  3|  feet.  Hence,  2} 
feet  and  3f  feet  are  multiples  of  ^  of  a  foot,  2|  feet  being  obtained 
by  taking  ^  of  a  foot  15  times,  and  3|  by  taking  ^  of  a  foot  22  times. 

327.  When  two  quantities  are  incommensurable,  that  is, 
have  no  common  unit  in  terms  of  w^hich  both  quantities  can 
be  expressed  in  integers,  it  is  impossible  to  find  a  fraction 
that  will  indicate  the  exact  value  of  the  ratio  of  the  given 
quantities.  It  is  possible,  however,  by  taking  the  unit  suf- 
ficiently small,  to  find  a  fraction  that  shall  differ  from  the 
true  value  of  the  ratio  by  as  iittle  as  we  please. 

Thus,  if  a  and  h  denote  the  diagonal  and  side  of  a  square, 

^=  V2. 
h 

.  Now  \/2  =  1.41421356 ,  a  value  greater  than  1.414213,  but  less 

than  1.414214. 

If,  then,  a  millionth  part  of  h  be  taken  as  the  unit,  the  value  of  the 
ratio  ^  lies  between  H^^f ^f  and  jf^^f^f ,  and  therefore  differs  from 
either  of  these  fractions  by  less  than  tooooott- 

By  carrying  the  decimal  farther,  a  fraction  may  be  found  that  will 
differ  from  the  true  value  of  the  ratio  by  less  than  a  billionth,  tril- 
lionth,  or  any  other  assigned  value  whatever. 

328.  Expressed  generally,  when  a  and  b  are  incommen- 
surable, and  b  is  divided  into  any  integral  number  (n)  of 
equal  parts,  if  one  of  these  parts  be  contained  in  a  more 
than  m,  times,  but  less  than  'm-j-l  times,  then 

a^m  7n-i-l  _ 

J  >  — ,  but  < ; 

on  n 

that  is,  the  value  of  -  lies  between  —  and  — ^I^. 
6  n  n 


^80  ALGEBRA. 


The  error,  therefore,  in  taking  either  of  these  values  for 

—  is  <  ~.  But  by  increasing  n  indefinitely,  -  can  be  made 
on  n 

to  decrease  indefinitely,  and  to  become  less  than  any  as- 
signed value,  however  small,  though  it  cannot  be  made 
absolutely  equal  to  zero. 

329.  The  ratio  between  two  incommensurable  quantities 
is  called  an  incommensurable  ratio. 

330.  As  the  treatment  of  Proportion  in  Algebra  depends 
upon  the  assumption  that  it  is  possible  to  find  fractions 
which  will  represent  the  ratios,  and  as  it  appears  that  no 
fraction  can  be  found  to  represent  the  exact  value  of  an 
incommensurable  ratio,  it  is  necessary  to  show  that  two 
incommensurable  ratios  are  eqiigil  if  their  true  values  always 
lie  between  the  same  limits,  however  little  these  limits  differ 
from  each  other. 

Let  a :  h  and  c:  dhe  two  incommensurable  ratios. 

Suppose  the  true  values  of  the  ratios  a  :  h  and  c  :  d  \\q  between 

V}l  and  ^  .  Then  the  difference  between  the  true  values  of  these 
n  n  ^  -■ 

ratios  is  less  than  -,  however  small  the  value  of  -  may  be.  ^  328. 

n  n 

But  since  -  can  be  made   to   approach   zero  at  pleasure,  _  can 
n  n 

be  made  less  than  any  assumed  difference  between  the  ratios. 

Therefore,  to  assume  any  difference  between  the  ratios  is  to  assume 

it  possible  to  find  a  quantity  that  for  the  same  value  of  —  shall  be 

1  ^* 

both  greater  and  less  than  _  ;  which  is  a  manifest  absurdity. 
n 
Hence,  a-.h  =  c-.d. 

331.  It  will  be  well  to  notice  that  the  word  limit  means 
a  fixed  value  from  which  another  and  variable  value  may 
be  made  to  differ  by  as  little  as  we  please ;  it  being  impos- 
sible, however,  for  the  difference  between  the  variable  value 
and  the  limit  to  become  absolutely  zero. 


RATIO.  2S1 


332.    A  ratio  will  not  be  altered  if  both  its  terms  be  multi- 
plied by  the  same  number. 

For  the  ratio  a :  &  is  represented  by  -,  the  ratio  ma :  rnh  is  repre- 
sented by  ^ ;  and  since  ^  =  ^,  .-.  ma  \m'h  =  a\'b. 
mo  mo      h 


333.  A  ratio  will  be  altered  if  different  multipliers  of  its 
terms  be  taken ;  and  will  be  increased  or  diminished  accord- 
ing as  the  multiplier  of  the  antecedent  is  greater  or  less  than 
that  of  the  consequent. 

For,      ma :  nh  will  be  >  or  <  a :  5 

according  as  !!^  is  >  or  <  ^  f  =  ^\ 

^  nh  h\     nbj 

as  ma  is  >  or  <  na, 

as  m    is  >  or  <  n. 

334.  A  ratio  of  greater  inequality  will  be  diminished,  and 
a  o'atio  of  less  inequality  increased  by  adding  the  same  nwm- 
ber  to  both  its  terms. 

For,  a+a;:5  +  a;is>or<a:5 

according  as  ^         is  >  or  <  -, 

h  +  X  h 

as        ah  -\-  hx  \s  >  or  <  ah  +  ax, 

as  hx\%>  or  <  ax, 

as  6  is  >  or  <  a. 

335.  A  ratio  of  greater  inequality  will  be  increased,  and  a 
ratio  of  less  inequality  diminished,  by  subtracting  the  same 
number  from  both  its  terins. 

For,  a  —  x:h—x  will  be  >  or  <  a :  6 

according  as  ^  ~^  is  >  or  <  -, 

h  —X  h 

as        ah  ~  hx  ia  >  or  <  ah  —  ax, 

as  ax  is  >  or  <  hx, 

as  a  is  >  or  <  &. 


282 


ALGEBRA. 


336.  Ratios  are  compounded  by  taking  the  product  of  the 
fractions  that  represent  them. 

Thus,  the  ratio  compounded  of  a:b  and  c:  d  is  found  by  taking  the 

product  of  ^  and  -  =  ^. 
b  d     hd 

The  ratio  compounded  of  a:b  and  a:  b  is  the  duplicate  ratio  a^ :  h^, 

and  the  ratio  compounded  of  o :  6,  a:b,  and  a:b  is  the  triplicate  ratio 

a^:b\ 

337.  Ratios  are  compared  by  comparing  the  fractions  that 
represent  them. 

Thus,  a:b  is  >  or  <  c:d 

according  as  -  is  >  or  <  — , 

^  b  d' 

as  ^is>or<-^, 

bd  bd 

as  ad  is  >  or  <  be. 


Exercise  CXIII. 

1.  Write   down  the  ratio  compounded  of  3  :  5  and  8  :  7. 

Which  of  these  ratios   is   increased,   and  which  is 
diminished  by  the  composition? 

2.  Compound  the  duplicate  ratio  of  4  :  15  with  the  tripli- 

cate of  5 :  2. 

3.  Show  that  a  duplicate  ratio  is  greater  or  less  than  its 

simple  ratio  according  as  it  is  a  ratio  of  greater  or 
less  inequality. 

4.  Arrange  in  order  of  magnitude  the  ratios  3:4;  23  :  25 ; 

10:11;  and  15: 16. 

5.  Arrange  in  order  of  magnitude 

a-\-'b:a  —  h  and  a^^h^  :o?  —  W,  if  a>h. 

Find  the  ratios  compounded  of: 

6.  3:5;   10:21;   14:15.  7.    7  :  9 ;  102:105;   15:17. 


1 


RATIO.  283 


8.  a^^ax-\'X^        and  ^^^~^^  +  ^. 

c?  —  c^x^ao^  —  o?  a-\-x 

9    -^^-9^  +  20  ^^^  3?~\Zx-\-^2 
ci?  —  Q>x  0(?  —  hx 

10.  a-\-h:a-h;  a^ -\- h"" -.{a -^  hf  ■  (a'~by:a'~b\ 

11.  Two  numbers  are  in  the  ratio  2  :  3,  and  if  9  be  added 

to  each,  they  are  in  the  ratio  3  :  4.     Find  the  num- 
bers. 
(Let  2  X  and  3  x  represent  the  numbers). 

12.  Show  that  the  ratio  a :  Z>  is  the  duplicate  of  the  ratio 

a-{-  c:h-\~c,  if  c^  =  ah. 

13.  Find  two  numbers  in  the  ratio  3:4,  of  which  the  sum 

'is  to  the  sum  of  their  squares  in  the  ratio  of  7  to  50. 

14.  If  five  gold  coins  and  four  silver  ones  be  worth  as  much 

as  three  gold  coins  and  twelve  silver  ones,  find  the 
ratio  of  the  value  of  a  gold  coin  to  that  of  a  silver  one. 

15.  If  eight  gold  and  nine  silver  coins  be  worth  as  much 

as  six  gold  and  nineteen  silver  coins,  find  the  ratio 
of  ths  value  of  a  silver  coin  to  that  of  a  gold  one. 

16.  There  are  two  roads  from  A  to  B,  one  of  them  14  miles 

longer  than  the  other ;  and  two  roads  from  B  to  C, 
one  of  them  8  miles  longer  than  the  other.  The  dis- 
tance from  A  to  B  is  to  the  distance  from  B  to  G,  by 
the  shorter  roads,  as  1  to  2 ;  by  the  longer  roads,  as 
2  to  3.     Find  the  distances. 

17.  What  must  be  added  to  each  of  the  terms  of  the  ratio 

m  :  n,  that  it  may  become  equal  to  the  ratio p  :  q'^ 

18.  A  rectangular  field  contains  5270  acres,  and  its  length 

is  to  its  breadth  in  the  ratio  of  31  :  17.  Find  its  di- 
mensions. 


284  ALGEBRA. 


Proportion. 


338.  An  equation  consisting  of  two  equal  ratios  is  called 
a  proportion ;  and  the  terms  of  the  ratios  are  called  propor- 
tionals. 

339.  The  algebraic  test  of  a  proportion  is  that  the  two 

fractions  which  represent  the  ratios  shall  be  equal. 

Thus,  the  ratio  a :  b  will  be  equal  to  the  ratio  c  :  dZ  if  y  =  -  ;  and 

0      d 
the  four  numbers  a,  b,  c,  d  are  called  proportionals,  or  are  said  to  be 

in  proportion. 

340.  If  the  ratios  a  :  b  and  c  :  d  form  a  proportion,  the 
proportion  is  written  -h—    •  1 

(read  the  ratio  of  a  to  6  is  equal  to  the  ratio  of  c  to  c?) 

or        '  a\h  :  :  c  :  d 

(read  a  is  to  b  in  the  same  ratio  as  c  is  to  d). 

The  first  and  last  terms,  a  and  d,  are  called  the  extremes. 

The  two  middle  terms,  b  and  c,  are  called  the  means. 

341.  When  four  numbers  are  in  proportion,  tJie  product 

of  the  extremes  is  equal  to  the  product  of  the  means. 

For,  if  a.b  ::  c:d, 

.-,  a      c 

By  multiplying  by  bd,   ad  =  be. 

342.  If  the  product  of  two  numbers  be  equal  to  the  product 
of  two  others,  either  tiuo  may  be  made  the  extremes  of  a  'pro- 
portion and  the  other  two  the  means. 

For,  if  ad  =  be, 

ad     be 


by  dividing  by  bd, 


b     ct 
a:  b  ::  e:  d. 


I 


PROPORTION.  285 


343.  The  equation  ad  =  he  gives 

he       ,       ad 

a  =  -T ;    0  =  — ; 

a  c 

so  tliat  an  extreme  may  be  found  bv  dividing  the  product 
of  the  means  by  the  other  extreme ;  and  a  mean  may  be 
found  by  dividing  the  product  of  the  extremes  by  the  other 
mean. 

344.  If  four  quantities,  a,  h,  c,  d,  be  in  proportion,  they 
will  be  in  proportion  by  : 

I.   Inversion. 

That  is,  h  will  be  to  a  as  c?  is  to  c. 

For,  if        a:h  ::  c:d, 


then 

h^d' 

and 

^      a      .      e 

or 

b  _d 

a      c* 

.-.b-.a  \:  die. 

345.   II. 

Composition. 

That  is, 

a-\-h  will  be  to  5  as  e  +  ^  is  to  d. 

For,  if 

a:b  .:  c:d. 

then 

a      c 

and 

a      -I 
5+1  = 

-h^' 

or 

a  +  6 

b     ~ 

c  +  d 
d    ' 

.a  +  b:b 

::C  +  d: 

d. 

346. 

III. 

Division. 

That 

is, 

a  —  h  will  be 

to  5  as  c  —  c?  is 

tod. 

For,  if 

a:b:: 

c:d. 

then 

a 

C 
'd' 

286  ALGEBRA. 


and  -  —  1  ==  -,  —  1 , 

b  d 

a  —  b      c  —  d 
.'.  a~b  :  b  :  :  C—  d:  d. 


347.    IV.    Composition  and  Division. 

That  is,  a-{-b  will  be  to  a  —  ^  as  c  +  c?  is  to  c~d. 

For,  from  II.,         «  +  ^      '  +  ^ 


an  J  from  III., 
By  dividing, 


b  d 

a  —  b  _ c  —  d 

b     ~     d    ' 
a  +  b  _^c  +  d 
a  —  b      c  —  d' 
a  -i-  b:  a—b  ::  c  \-  d:  c  —  d. 


348.    "When  the  four  quantities  a,  b,  c,  d  are  all  of  the 
sa'me  kmd,  they  will  be  iii  proportion  by  : 


V  .    .0.1  ucr. 

That  is, 

UUibXUlI. 

a  will  be  to  c  as  b  is  to  d. 

For,  if                             a.b  ::  c:d, 

then                                     «      |. 

b       d 

By  multiplying  by  ^,     f  =  ^, 
c       be      cd 

"4' 

c      d 

.-.  a:  c  ::  b.d. 

349.    From  the  proportion  a:  c  : :  b  :  d  may  be  obtained 
by: 

VI.  Composition,    a  ~{-  c  :  c  : :  h  -\~  d :  d. 

VII.  Division.  a  —  c  :  c  : :  b  —  d :  d. 

VIII.  Composition  and  Division.  a-\-c:a  —  c::b-\-d:b  —  d. 


PROPOKTION.  287 


350.  In  a  series  of  equal  ratios,  the  sum  of  the  antecedents 
is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its 
consequent. 

For,  if  ^_^  =  1^£, 

b      d     f     h 

r  may  be  put  for  each  of  these  ratios. 

Then    —  =  r,  —  =  r,  -  =  r,  ^  =  r, 
b  d  f  h 

.'.  a=br,  c  =  dr,  e  =fr,  g  =  hr. 

.■ .  a  +  c  +  e  +  g  =  {b  +  d  +f  +  h)  r. 

.  a  +  c  +  e  +  g  _    _a 

b  +  d+f+h  ~^~b' 

.\a  +  c  +  e  +  g  :  b  +  d  +/+  h  ■ :  a  :  b. 

In  like  manner,  it  may  be  shown  that 

ma  +  710  +pe  +  qg  :  mb  +  nd  +pf  +  qh  ::  a  :  b. 

351.  li  a,  b,  c,  d  be  in  continued  proportion,  that  is,  if 
a'.h=^b  :  c=c:  d,  then  will  a:  c-=a^  \h^  and  a:  d=a^ :  h^. 

For, 

II  en  CO, 


So 


a      b      c 

b~  c~  d 

b^  c^b^F 

a      a* 

c      ¥' 

.■.a:c=^a^:b^ 

a 
b 

b      c      a      a      a 

a      a? 

d      l^' 

.■.a:d=a^:b\ 

352.  If  a,  b,  c  be  proportionals,  so  that  a:.b  wb  :  c,  then 
b  is  called  a  mean  proportional  between  a  and  c,  and  c  is 
called  a  third  proportional  to  a  and  b. 


288  ALGEBRA. 

If  a : , 


h  :  c,  then  h  = 

=  -\/ac. 

For,  if 

a:b  ::  b:c. 

then 

a      b 
b       c' 

and 

b''  =  ac, 

:.  b  =  Vac 

353.    The  products  of  the  corresponding  terms  of  two  or 
more  proportions  are  in  proportion. 


For,  if 

a:b  ::  c:d, 

e:f::g:h. 

and 

h:l ::  m:n, 

then            «. 

c      e      g     h      m 
~d'   f~h'    l~n' 

Hence,  by  finding  the  product  of  the  left  members,  and  also  of  the 
right  members  of  these  equations, 

aek  _  cgrn 

bfl     dhn 

.'.  aek :  bfl : :  cgm :  dhri. 

354.    Idke  powers,  or  like  roots,  of  the  terms  of  a  propor- 
tion are  in  proportion. 

For,  if  a:b  ::  c:d, 

iu  a      c 

then  ^  =  -. 

By  raising  both  sides  to  the  nth  power, 

.'.  a*» :  6"  :  :  C* :  d^. 
By  extracting  the  nth  root, 


PROPORTION.  289 


355.  Ij  two  quantities  he  increased  or  diminished  by  like 
parts  of  each,  the  results  will  he  in  the  same  ratio  as  the 
quantities  themselves. 


For. 


a 


that  is. 


■.  a:  b 


356.  The  laws  that  have  been  established  for  ratios 
should  be  rememeinbered  when  ratios  are  expressed  in 
their  fractional  form. 

cc'  +  x  +  l      x'-x  +  2 


a^  —  x—\      a^-\-x  —  2 
2x'  2ar» 


(1)  Solve: 

By  I  347 

and  this  equation  is  satisfied,  when  a;  =  0 ; 

A-   -A-      V    2r»  11 

or,  dividing  by  -—, =  ——, 

2  x+1     2—x 

.'.  x  =  ^. 

[2)  li  a  :  h  : :  c  :  d,  show  that 

a^  +  ab-.P-ah::  c" -\^  cd :  d'' ~  cd. 

If 

then  ^  =  :-^.  ^347. 

and 


a  _ 
b~ 

c 
"d' 

a  +  b 
a-b 

c  +  d 
e-d' 

a 
-b 

e 

b 

^a  +  b 

-^a- 

C  + 

c  — 

d' 

353. 


that  is. 


a^  -\-  ab  ^  (?  +  cd 

b^-ab      d^-cd 

(^  +  ab  :  l^  —  ab  :  :  c^  +  cd :  d^  —  cd. 


290  ALGEBRA. 


(3)   When  a  \  b  \ :  c  :  d,  and  a  is  the  greatest  term,,  show  that 
a  +  c?  is  greater  than  h-^-  c. 

Since  ?.  =  ^,  and  a  >  c, 


h 


.-.  h>d. 


Also,  since  '      ^-^  =  ^^,  ^46. 

h  d 

and  h  >  d, 

.-.  a  ~  b>  c  —  d. 

By  adding,  b  +  d  ==  h  +  d, 

a  +  d>  b  +  c. 


Exercise  CXIV. 
li  a  :  h  : :  c  :  d,  prove  that : 

1.  ma  :nh  ::  mc  :  nd.  4.    o^  -.W  w  c^  :  d^. 

2.  2>a-^h  :h  •.:oc-{-d:d.  5.    a:  a-}-})  :  :  c  :  c-{- d. 

3.  a-{-2h  -.b  \:  c-\-2d:  d.  6.    a  :  a  —  b  :  \  c  :  c  —  d. 

7.  ma  +  wZ) :  7)ia  —  nb::  mc  -f-  nd :  mc  —  72c?. 

8.  2a  +  Sb:3a-4:b  ■.:2c  +  Sd:Sc-4:d. 

9.  ma^  +  n(^  :  mb^  +  ^'^^^  :  :  a^  :  b^. 

10.    ma^  +  nab  -i-pb^  :  m.<?  +  '^^^  -{-pd^  :  :  b^ :  d^. 

If  a  :  b  :  :  b  :  c,  prove  that : 
11.    a  +  b  :b-}-c  ::  a:b.  12.    a^ +  ab  :  b^ -i-bc  :  :  a:  c. 

13.    a:c  ::  (a  +  bf  :  (b -\- cf. 

14.  When  a,  5,  and  c  are  proportionals,  and  a  the  greatest, 

show  that  a-{-c>  2b. 

15.  lf^ZL^  =  3^Zl^  =  ^IZ^,  and  :r,  y,  z  be  unequal,  then 

^  m  n 

Z  +  m  +  w  =  0. 


PROPORTION.  291 


16.  Find  X  when  x-\-b  :2x—?>  :  :  bx-^l  :?>x  —  ^. 

17.  Find  x  when  x-\-a:2x  —  h  :  :  ox-\-b  -.^iX—- a. 

18.  Find  X  when  -^/x  +  V6  :  V.r  —  V6  :  :  a:b. 

19.  Find  a;  and  y  when  a; :  27  ;  :  y  :  9,  and  a; :  27  :  :  2  :  x—t/. 

20.  Find  a;  and  y  when  a;  +  y+l:^  +  y  +  2::  6:7,  and 

when  y-]-2x  :^  —  2x  :  :  12:r  +  6y  — 3  :  6y— 12:r  — 1. 

21.  Find  x  when  a;^— 4a;4-2  :  a^-2x-l  : :  a:2_4^  .  ci^-2x-2. 

22.  A  railway  passenger  observes  that  a  train  passes  him, 

moving  in  the  opposite  direction,  in  2  seconds ;  but 
moving  in  the  same  direction  with  him,  it  passes  him 
in  30  seconds.     Compare  the  rates  of  the  two  trains. 

23.  A  and  B  trade  with  different  sums.     A  gains  $200  and 

B  loses  $50,  and  now  A's  stock  :  B's  :  :  2:  h  But,  if 
A  had  gained  $100  and  B  lost  $85,  their  stocks 
would  have  been  as  15  :  3i.  Find  the  original  stock 
of  each. 

24.  A  quantity  of  milk  is  increased  by  watering  in  the  ra- 

tio 4  •  5,  and  then  3  gallons  are  sold  ;  the  remainder 
is  mixed  with  3  quarts  of  water,  and  is  increased  in 
the  ratio  6  :  7.  How  many  gallons  of  milk  were  there 
at  first? 

25.  In  a  mile  race  between  a  bicycle  and  a  tricycle  their 

rates  were  as  5  :  4.  The  tricycle  had  half  a  minute 
start,  but  was  beaten  by  176  yards.  Find  the  rates 
of  each. 

26.  The  time  which  an  express- train  takes  to  travel  180 

miles  is  to  that  taken  by  an  ordinary  train  as  9  :  14. 
The  ordinary  train  loses  as  much  time  from  stopping 
as  it  would  take  to  travel  30  miles ;  the  express-train 
loses  only  half  as  much  time  as  the  other  by  stopping, 
and  travels  15  miles  an  hour  faster.  What  are  their 
respective  rates? 


292  ALGEBRA. 


27.  A  line  is  divided  into  two  parts  in  the  ratio  2  :  3,  and 

into  two  parts  in  the  ratio  3:4;  the  distance  be- 
tween the  points  of  section  is  2.  Find  the  length  of 
the  line. 

28.  A  railway  consists  of  two    sections ;    the  annual    ex- 

penditure on  one  is  increased  this  year  5%,  and  on 
•     the  other  4%,  producing  on  the  whole  an  increase  of 
4-3^%.     Compare  the  amount  expended  on  the  tw^o 
sections  last  year,  and  also  this  year. 

29.  When  a,h,c,d  are  proportional  and  unequal,  show  that 

no  number  x  can  be  found  such  that  a-TX,  h-{-x, 
c-\-x,  d-i-x  shall  be  proportionals. 


Variation. 

357.  Two  quantities  may  be  so  related  that,  when  one  has 
its  value  changed,  the  other  will,  in  consequence,  have  its 
value  changed. 

Thus,  the  distance  travelled  in  a  certain  time  will  be  doubled  if 
the  rate  be  doubled.  The  time  required  for  doing  a  certain  quantity 
of  work  will  be  doubled  if  only  half  the  number  of  workmen  be 
employed. 

358.  Whenever  it  becomes  necessary  to  express  the  c/en- 
eral  relations  of  certain  kinds  of  quantities  to  each  other, 
without  confining  the  inquiry  to  any  particular  values  of 
these  quantities,  it  will  usually  be  sufficient  to  mention  two 
of  the  terms  of  a  proportion.  In  all  such  cases,  however, 
four  terms  are  always  implied. 

Tlius,  if  it  be  said  that  the  weight  of  water  is  proportional  to  its 
volume,  or  varies  as  its  volume,  the  meaning  is,  that  one  gallon  of 
water  is  to  any  number  of  gallons  as  the  weight  of  one  gallon  is  to  the 
weight  of  the  given  number  of  gallons. 


VARIATION.  293 


359.  Quantities  used  in  a  general  sense,  as  distance,  time, 
weight,  volume,  to  which  particular  values  may  be  assigned, 
are  denoted  by  capital  letters.  A,  B,  C,  etc. ;  while  as- 
signed values  of  these  quantities  may  be  denoted  by  small 
letters,  a,  b,  c,  etc.  The  letters  A,  B,  Cwill  be  understood 
to  represent  any  nuTuerical  values  that  may  be  assigned  to 
the  quantities ;  and  when  two  such  letters  occur  in  an  ex- 
pression they  will  be  understood  to  represent  any  corre- 
sponding numerical  values  that  may  be  assigned  to  the  two 
quantities. 

360.  When  two  quantities  A  and  B  are  so  connected 
that  their  ratio  is  constant,  that  is,  remains  the  same  for  all 
corresponding  values  of  A  and  B,  the  one  is  said  to  vary  as 
the  other ;  and  this  relation  is  expressed  by  ^  oc  ^  (read 
A  varies  as  B). 

Thus,  the  area  of  a  triangle  with  a  given  base  varies  as  its  altitude ; 
for,  if  the  altitude  be  changed,  the  area  will  be  changed  in  the  same 
ratio. 

A 

If  this  constant  ratio  be  denoted  by  m,  then  —  =  m,  or  ^ 

=  mB.  ^ 

From  this  equation  m  may  be  found  when  two  corre- 
sponding values  of  A  and  B  are  known, 

361.  When  two  quantities  are  so  connected  that  if  one 
be  changed  in  any  ratio,  the  other  will  be  changed  in  the 
inverse  ratio,  the  one  is  said  to  vary  inversely  as  the  other. 

Thus,  the  time  required  to  do  a  certain  amount  of  work  varies  in- 
versely as  the  number  of  workmen  employed  ;  for,  if  the  number  of 
workmen  be  doubled,  halved,  or  changed  in  any  ratio,  the  time  re- 
quired will  be  halved,  doubled,  or  changed  in  the  inverse  ratio. 

362.  If  A  vary  inversely  as  B,  two  values  of  A  have  to 
each  other  the  inverse  ratio  of  the  two  corresponding  values 
of  ^  ;  or  a:  a'  ::h'  :b\  that  is,  ab  =  a'b'. 


294  ALGEBEA. 

Hence,  the  product  AJB  is  constant,  and  may  be  denoted 
by  m.     That  is,  AB  =  m. 

If  any  two  corresponding  values  of  A  and  B  be  known, 
the  constant  m  may  be  found. 

The  equation  A£  =  m  may  be  written  A=—,  and  as  m 
is  constant,  A  is  said  to  vary  as  the  reciprocal  of  B,  or 

363.  The  two  equations, 

A  =  mB  (for  direct  variation), 

^  =  -—  (for  inverse  variation), 
B 

furnish  the  simplest  method  of  treating  Variation. 

If  A  =  mBC,  A  is  said  to  y&yj  jointli/  as  B  and  C. 

If  ^  =  -77-,  A  is  said  to  vary  directly  as  B  and  inversely 
as  (7. 

364.  The  following  results  are  to  be  observed : 

I.  If  J.  oc  ^  and  .5  oc  C,  then  A  a:  C. 

For  A  ==  mB,  where  m  is  constant, 

and  B  =  nC,    where  n  is  constant. 

.-.  A  =  mnC. 
.'.  A  oc  C,  since  mn  is  constant. 

In  like  manner,  \i  A  cc  B  and  B  cc  —,  then  A  k  —. 

II.  If  ^  oc  (7 and  ^  oc  C,  then  ^  ±  ^  a  C,  and  VXB  oc  C. 

For  -4  =  m(7,  where  rn-is  constant, 

and  B  =  nC,  where  7i  is  constant. 

.•.A±B=^{m±n)  C. 

:.  A  ±  B  cc  C,  since  m  ±  n  is  constant. 
Also,     y/AB=  V'mCxnC=  VmnG^=  CVmn. 

.•.  VAB  oc  C,  since  Vwm  is  constant. 


VARIATION.  295 


III.    li  A  a:  B&ndCocI),  then  AOo:£I). 

For  .  A  =  mB,  where  m  is  constant, 

C=nl),    where /I  is  constant. 
.■.AC=  mnBD. 
.'.  AC ^  BD,  since  mn  is  constant. 


IV.    liAazB  then  A""  oc  ^'». 

For  A  =  mB,  where  m  is  constant. 

.'.  A*  =  m^B*^. 
.'.  A*  oc  B**,    since  m*  is  constant 

Y.  If  ^  cc  ^  when  C  is  unchanged,  and  A  cc  C  wh«n  i? 
is  unchanged,  then  A  oc  ^Cwhen  both  B  and  (7  change. 

For         A  =  mB,  when  5  varies  and  C  is  constant. 
Here,  m  is  constant  and  cannot  contain  th«  variable  B 
.'.  A  must  contain  B^  but  no  other  power  of  B. 
Again,     A  =  wC,  when  C  varies  and  B  is  constant. 
Here,  n  is  constant  and  cannot  contain  the  variable  C^ 
.'.  ^  must  contain  C,  but  no  other  power  of  C. 
Hence,  A  contains  both  B  and  C,  but  no  other  powers  of  B 
and  C,  and  therefore, 

— -  =p,  or  A  =pBC,  where  _p  is  constant 

.*,  J.  oc  BC,    since  p  is  constant. 

In  like  manner,  it  may  be  shown  that  if  A  vary  as  each 
of  any  number  of  quantities  B,  C,  I),  etc.,  when  the  rest 
are  unchanged,  then  when  they  all  change,  A  oc  BCD,  etc. 

Thus,  the  area  of  a  rectangle  varies  as  the  base  when  the  altitude 
IS  constant,  and  as  the  altitude  when  the  base  is  constant,  but  as  the 
product  of  the  base  and  altitude  when  both  vary. 

The  volume  of  a  rectangular  solid  varies  as  the  length  when  the 
width  and  thickness  remain  constant ;  as  the  width  when  the  length 
and  thickness  remain  constant ;  as  the  thickness  when  the  length  and 
width  remain  constant;  but  as  the  product  of  length,  breadth,  and 
thickness  when  all  three  vary. 


296  ALGEBEA. 


(1)  If  A  vary  inversely  as  B,  and  when  A  =  2  the  corre- 

sponding value  of  B  is  36,  find  the  corresponding 
value  of  JB  when  ^  =  9. 

Here  ^  =  ^, 

or  m  =  AB, 

.•.m  =  2x36  =  72. 
And  if  9  and  72  be  substituted  for  A  and  m  respectively  in 

72 
the  result  is         9  =  — , 
B 

.-.9.5=72. 

.•.B=S.  Ans. 

(2)  The  weight  of  a  sphere  of  given  material  varies  as  its 

volume,  and  its  volume  varies  as  the  cube  of  its  diam- 
eter. If  a  sphere  4  inches  in  diameter  weigh  20 
pounds,  find  the  weight  of  a  sphere  5  inches  in  diam- 
eter. 

Let     TF"  represent  the  weight, 
V  represent  the  volume, 
D  represent  the  diameter. 
Then  TFocFand  Foe  D^, 

.-.  Woe  D^. 
Put      W=  mD^, 
then,  since  20  and  4  are  corresponding  volumes  of  TFand  D, 
20  =  771  X  64, 

.:  when  D  =  5,  TF=  j\  of  125  =  39^. 

Exercise  CXV. 

1.  If  J[  oc  B,  and  A  =  4:  when  B  =  5,  find  A  when  B  =  12. 

2.  If^oc  ^,  and  when  B  =  i,  A  =  i,  find  ^when^  =  i 

3.  If  A  vary  jointly  as  B  and  C,  and  3,  4,  5  be  simulta- 

neous values  of  A,  B,  C,  find  A  when  B  =  C—  10. 


I 


VARIATION.  297 


4.  If  ^  oc  — ,  and  when  A  =  10,  B  =  2,  find  the  value  of 

B  when  A  =  4:. 

5.  li  Ace  — ,  and  when  ^  =  6,  ^  =  4,  and   C=  3,  find 

G 
the  value  of  A  when  J3  =  5  and  0=7. 

6.  If  the  square  of  JTvary  as  the  cube  of  Y,  and  Jr=  3 

when  Y—  4,  find  the  equation  between  Xand  Y. 

7.  If  the  square  of  X  vary  inversely  as  the  cube  of  Y,  and 

X=  2  when  y=  3,  find  the  equation  between  X 
and  Y. 

8.  If  Z  vary  as  X  directly  and  Y  inversely,  and  if  when 

Z=2,  X=3,  and  r=4,  find  the  value  of  ^  when 
X=  15  and  Y=  8. 

9.  li  A  cc  B  -\~  c  where  c  is  constant,  and  if  ^  =  2  when  B 

=  1,  and  if  ^  =  5  when  B  ^2,  find  A  when  B  =  3. 

10.  The  velocity  acquired  by  a  stone  falling  from  rest  varies 

as  the  time  of  falling;  and  the  distance  fallen  varies 
as  the  square  of  the  time.  If  it  be  found  that  in  3 
seconds  a  stone  has  fallen  145  feet,  and  acquired  a 
velocity  of  961  feet  per  second,  find  the  velocity  and 
distance  at  the  end  of  5  seconds. 

11.  If  a  heavier  weight  draw  up  a  lighter  one  by  means  of 

a  string  passing  over  a  fixed  wheel,  the  space  de- 
scribed in  a  given  time  will  vary  directly  as  the 
difierence  between  the  weights,  and  inversely  as  their 
sum.  If  9  ounces  draw  7  ounces  through  8  feet  in  2 
seconds,  how  high  will  12  ounces  draw  9  ounces  in 
the  same  time  ? 

12.  The  space  will  vary  also  as  the  square  of  the  time. 

Find  the  space  in  Example  11,  if  the  time  in  the  lat- 
ter case  be  3  seconds. 


298  ALGEBRA. 

13.  Equal  volumes  of  iron  and  copper  are  found  to  weigh 

77  and  89  ounces  respectively.  Find  the  weight  of 
10  J  feet  of  round  copper  rod  when  9  inches  of  iron 
rod  of  the  same  diameter  weigh  31y^^  ounces. 

14.  The  square  of  the  time  of  a  planet's  revolution  varies  as 

the  cube  of  its  distance  from  the  sun.  The  distances 
of  the  Earth  and  Mercury  from  the  sun  being  91  and 
35  millions  of  miles,  find  in  days  the  time  of  Mer- 
cury's revolution. 

15.  A  spherical  iron  shell  1  foot  in  diameter  weighs  -^^  of 

what  it  would  weigh  if  solid.  Find  the  thickness 
of  the  metal,  it  being  known  that  the  volume  of  a 
sphere  varies  as  the  cube  of  its  diameter. 

16.  The  volume  of  a  sphere  varies  as  the  cube  of  its  diame- 

ter. Compare  the  volume  of  a  sphere  6  inches  in 
diameter  with  th^um  of  the  volumes  of  three  spheres 
whose  diameters  are  3,  4,  5  inches  respectively. 

17.  Two  circular  gold  plates,  each  an  inch  thick,  the  diam- 

eters of  which  are  6  inches  and  8  inches  respectively, 
are  melted  and  formed  into  a  single  circular  plate 
1  inch  thick.  Find  its  diameter,  having  given  that 
the  area  of  a  circle  varies  as  the  square  of  its  diameter. 

18.  The  volume  of  a  pyramid  varies  jointly  as  the  area  of 

its  base  and  its  altitude.  A  pyramid,  the  base  of  which 
is  9  feet  square,  and  the  height  of  which  is  10  feet,  is 
found  to  contain  10  cubic  yards.  What  must  bo  the 
height  of  a  pyramid  upon  a  base  3  feet  square,  in 
order  that  it  may  contain  2  cubic  yards  ? 


CHAPTER  XXI. 
Series. 

365.  A  succession  of  numbers  which  proceed  according 
to  some  fixed  law  is  called  a  series;  and  the  successive 
numbers  are  called  the  terms  of  the  series. 

Thus,  by  executing  the  indicated  division  of ,  the  series  1  +x 

1—x 

+  x^  +  'S(^  + is  obtained,  a  series  that  has  an  unlimited  number  of 

terms. 

366.  A  series  that  is  continued  indefinitely  is  called  an 
infinite  series ;  and  a  series  that  comes  to  an  end  at  some 
particular  term  is  called  a  finite  series. 

367.  "When  a:  is  <  1,  the  more  terms  we  take  of  the  infi- 
nite series  l  +  a7-f-^  +  ^+ ,  obtained  by  dividing  1  by 

1—x,  the  more  nearly  does  their  sum  approach  to  the  value 

\—x 

1  13 

Thus,  if  a;  =  ^  then = •  =  -,  and  the  series  becomes  1  +  J 

1—x      1—^      2 

+  i  +  iV  + '  ^  ^^^  which  cannot  become  equal  to  f  however  great 

the  number  of  terms  taken,  but  which  may  be  made  to  differ  from  f 
by  as  little  as  we  please  by  increasing  indefinitely  the  number  of 
terms. 

368.  But  when  x  is  >  1,  the  more  terms  we  take  of  the 

series  l-^-x-^D^-^-cf^-^- the  more  does  the  sum  of  the 

series  diverge  from  the  value  of 


1-^ 

2'' 
3  +  9  +  27  + ,  a  sum  wliich   diverges  more   and  mor-e   from  —  J, 


Thus,  if  a;  =  3,  then = ■  = ,  and  the  series  becomes  1  + 

1  -  .T      1-3  2 


300  ALGEBRA. 


the  more  terms  we  take,  and  which  may  be  made  to  increase  indefi- 
nitely by  increasing  indefinitely  the  number  of  terms  taken. 

369.  A  series  whose  sum  as  the  number  of  its  terms  is  in- 
definitely increased  approaches  some  fixed  finite  value  as  a 
limit  is  called  a  converging  series ;  and  a  series  whose  sum 
increases  indefinitely  as  the  number  of  its  terms  is  increased, 
is  called  a  diverging  series. 

370.  When  x^=l,  the  division  of  1  by  1  — :r,  that  is,  of 
1  by  0,  has  no  meaning,  according  to  the  definition  of  divi- 
sion; and  any  attempt  to  divide  by  a  divisor  that  is  equal 
to  zero  leads  to  absurd  results. 

Thus,  8  +  4  =  8  +  4; 

by  transposing,  8  —  8  =  4  —  4; 

or,  dividing  by  4  —  4,        2  =  1;         a  manifest  absurdity. 

371.  When  x=^l  very  nearly,  then  the  value  of 


will  be  very  great,  and  the  sum  of  the  series  1  +  a;  +  ^^  + 

x^-\- will  become  greater  and  greater  the  more  terms  we 

take.     Hence,  by  making  the  denominator  \—x  approach 
indefinitely  to  zero,  the  value  of  the  fraction may  be 

J.        X 

made  to  increase  at  pleasure. 

372.  If  the  symbol  o  be  used  to  denote  a  quantity  that 
is  less  than  any  assignable  quantity,  and  that  may  be  con- 
sidered to  decrease  without  limit,  not,  however,  becoming 
0,  and  the  symbol  Q?  be  used  to  denote  a  quantity  that  is 
greater  than  any  assignable  quantity,  and  that  may  be  con- 
sidered to  increase  without  limit,  not,  however,  becoming  oo, 
then  1 

_r=00 
O  '~' 

In  the  same  sense  ^  =  5?,  where  a  represents  any  value 
that  may  be  assigned. 


SERIES.  301 


I—  ^ 

373.  If  X  in  the  fraction  :; be  equal  to  1,  the  numer- 

1  —X 

ator  and  denominator  will  each  become  0,  and  the  fraction 
will  assume  the  form  -. 

374.  If,  however,  x  in  this  fraction  approach  to  1  as  its 
limit,  then  the  denominator  1  —  x,  inasmuch  as  it  has  sorne 
value,  even  though  less  than  any  assignable  value,  may  be 
used  as  a  divisor,  and  the  result  is  1  -}-  x  -\-  x^  -{-  x^  -^  x*. 
Hence,  it  is  evident  that  though  both  terms  of  the  fraction 
become  smaller  and  smaller  as  1  —  a:  approaches  to  0,  still 
the  numerator  becomes  more  and  more  nearly  five  times  the 
denominator. 

It  may  be  remarked  that  when  the  symbol  %  is  obtained  for  the 
value  of  the  unknown  quantity  in  a  problem,  the  meaning  is  that  the 
problem  has  no  definite  solution,  but  that  its  conditions  are  satisfied 
if  any  value  whatever  be  taken  for  the  required  quantity  ;  and  if  the 
symbol  §,  in  which  a  denotes  any  assigned  value,  be  obtained  for 
the  value  of  the  unknown  quantity,  the  meaning  is  that  the  condi- 
tions of  the  problem  are  impossible. 

375.  The  number  of  different  series  is  unlimited,  but  the 
only  kinds  of  series  that  will  be  considered  at  this  stage  of 
the  work  are  Arithmetical,  Geometrical,  and  Harmonical 
Series. 

Arithmetical  Series. 

376.  A  series  in  which  the  difference  between  any  two 
adjacent  terms  is  equal  to  the  difference  between  any  other 
two  adjacent  terms,  is  called  an  Arithmetical  Series  or  an 
Arithmetical  Progression. 

377.  The  general  representative  of  such  a  series  will  be 

a,  a-\-d,  a-\-2d,  a-\-Sd...... 

in  which  a  is  the  first  term  and  d  the  common  difference ; 


302 


ALGEBRA. 


and  the  series  will  be  increasing  or  decreasing  according  as 
d  is  positive  or  negative. 

378.  Since  each  succeeding  term  of  the  series  is  obtained 
by  adding  d  to  the  preceding  term,  the  coefficient  of  d  will 
always  be  1  less  than  the  number  of  the  term,  so  that  the 

nth  term  =^  a '\- {n  —  1)  d. 

If  the  nth  term  be  denoted  by  I,  this  equation  becomes 

l^a-\-{n-l)d.  (1) 

379.  The  arithmetical  'mean  between  two  numbers  is  the 
number  which  stands  between  them,  and  makes  with  them 
an  arithmetical  series. 

380.  If  a  and  h  denote  two  numbers,  and  A  their  arith- 
metical mean,  then,  by  the   definition  of  an  arithmetical 

«^^^^^^  A-a  =  h-A, 

.■.A  =  ^^.  (2) 

2 

381.  Sometimes  it  is  required  to  insert  several  arithmeti- 
cal means  between  two  numbers. 

If  ?n  =  the  number  of  means,  then  m-{-2  =  n,  the  whole 
number  of  terms ;  and  if  m  +  2  be  substituted  for  n  in  the 
equation  l=a  +  (n-l)d, 

the  result  is  Z  =  a  -f  (m  -f  1)  c?. 

By  transposing  a,     l~a  =  (m-]-!)  d, 

Thus,  if  it  be  required  to  insert  six  means  between  3  and  17,  the 

17  —  3 
value  of  d  is  found  to  be =  2  ;  and  the  series  will  be  3,  5,  7,  9, 

6  +  1 
11,  13,  15,  17. 


SERIES.  303 

352.  If  I  denote  tlie  last  term,  a  the  first  term,  n  the 
number  of  terms,  d  the  common  difference,  and  s  the  sum 
of  the  terms,  it  is  evident  that 

s=      a    +(a+^)  +  («  +  2c?)  + -\-{l-d)-\-      I,  or 

_s  =      I     -{-(l-d)+(l-2d)  + +  (a-{-d)+      a 

.'.2s  =  (a+l)  +  (a-{-l)  +  (a+l)     + +  (a+0  +  (^+0 

—■n(a-\- 1) 

.■.s  =  l(a  +  r).  (4) 

383.    From  the  two  equations, 

l  =  a  +  (n~l)d,  (1) 

s  =  l(a+l),  (2) 

any  one  of  the  quantities  a,  d,  I,  n,  s  may  be  found  when 
three  are  given. 


Ex.  Find  n  when  d,  I,  s  are  given. 

From  (1),                 a  =  l-in-l)d. 

-r.         /ox                         2s  —  ln 

From  (2),                 a  = 

n 

Therefore,      I -(n-1)  d -^^~^^, 

.■.ln  —  dn^  +  dn  =  2s  —  ln, 

.\drv'-{2l  +  d)n  =  -2s. 

.'Ad^rv'-{  )  +  {2l  +  df  =  {21  +  df  -8ds, 

.:2dn-{2l  +  d)  =  ±  V(2 l  +  df-8 ds, 

2l  +  d±^{2l  +  df- 
2d 

-8ds 

Note.  The  table  on  the  following  page  contains  the  results  of  the 
general  solution  of  all  possible  problems  in  arithmetical  series.  The 
student  is  advised  to  work  these  out,  both  for  the  results  obtained 
and  for  the  practice  gained  in  solving  literal  equations  in  which  the 
unknown  quantities  are  represented  by  other  letters  than  x,  y,  z. 


304 


ALGEBRA. 


No. 

Given. 

Required, 

—I 

Results. 

1 
2 
3 
4 

a  d  n 
a  d  s 
a  n  s 
d  n  s 

I 

l=a  +  {n-l)d. 

l==-^d±^/[2ds  +  {a~idf\ 

r     (n-1)^ 
n^        2       ■ 

5 
6 

7 
8 

a  d  n 
a  d  I 
a  n  I 
dn  I 

s 

s  =  ^n[2a  +  {n-l)d]. 

2    '    2<; 

.=  (Z  +  a)|. 

s  =  in[2l-{n-l)d]. 

9 

10 
11 
12 

dn  I 
d  n  s 
d  I  s 
n  I  s 

a 

a  =  l-{n-l)d. 

a-'      (n-1)^ 
n            2       ' 

a  =  id±V{l  +  hdf-2ds. 

13 
14 
15 
16 

17 
18 
19 
20 

a  n  I 
a  n  s 
a  I  s 
n  I  s 

d 

n  —  1 

d     2(s-an) 
n{n-l) 

2s-l-a 
d     2(nZ-s) 
n{n-l) 

a  d  I 
ads 
a  I  s 
d  I  s 

n 

.=i^«... 

d-2a±V(2a~df  ridi 

2d 

l  +  a 

2l  +  d±  V{21  +  df-8d3 
2d 

SEKIES.  305 


Exercise  CXVI. 

1.  Find  the  thirteenth  term  of  5,  9,  13 

ninth  term  of  —  3,  —  1,  1 

tenth  term  of  —  2,  —  5,  —  8 

eighth  term  of  a,  a  +  3  5,  a -{-6  h 

fifteenth  term  of  1,  -f-,  -f- 

thirteenth  term  of —  48,  —44,  —40 

2.  The  first  term  of  an  arithmetical  series  is  3,  the  thir- 

teenth term  is  55.     Find  the  common  difierence. 

3.  Find  the  arithmetical  mean  between :    (a.)  3  and  12 ; 

{b.)  -  5  and  17  ;  (c.)  o? -\- ab  -  5^  and  a'-ab^  b\ 

4.  Insert  three  arithmetical  means  between  1  and  19;  and 

four  means  between  — 4  and  17. 

5.  The  first  term  of  a  series  is  2,  and  the  common  differ- 

ence i.     What  term  will  be  10  ? 

6.  The  seventh  term  of  a  series,  whose  common  difference 

is  3,  is  11.     Find  the  first  term, 

7.  Find  the  sum  of 

5  +  8  +  11  + to  ten  terms. 

—  4  —  1  +  2  + to  seven  terms. 

a  +  4a+  7a  + to  n  terms. 

-|  +  3^  +  ^  + to  twenty-one  terms. 

1  +  2|  +  4|-  + to  twenty  terms. 

8.  The  sum  of  six  numbers  of  an  arithmetical  series  is  27, 

and  the  first  term  is  1.     Determine  the  series. 

9.  How  many  terms  of  the  series  —5  —  2  +  1+ must 

be  taken  so  that  their  sum  may  be  63. 

10.    The  first  term  is  12,  and  the  sum  of  ten  terms  is  10. 
Find  the  last  term. 


306  ALGEBRA. 

11.  The  arithmetical  mean  between  two  numbers  is  10,  and 

the  mean  between  the  double  of  the  first  and  the 
triple  of  the  second  is  27.     Find  the  numbers. 

12.  Find  the  middle  term, of  eleven  terms  whose  sum  is  66. 

13.  The  first  term  of  an  arithmetical  series  is  2,  the  common 

difference  is  7,  and  the  last  term  79.  Find  the  num- 
ber of  terms. 

14.  The  sum  of  fifteen  terms  of  an  arithmetical  series  is  600, 

and  the  common  difference  is  5.     Find  the  first  term. 

15.  Insert  ten  arithmetical  means  between  —  7  and  114. 

16.  The  sum  of  three  numbers  in  arithmetical  progression 

is  15,  and  the  sum  of  their  squares  is  83.     Find  the 

numbers. 

Let  x  —  y,  X,  X  -^-y  represent  the  numbers. 

17.  Arithmetical  means  are  inserted  between  5  and  23,  so 

that  the  sum  of  the  first  two  is  to  the  sum  of  the  last 
two  as  2  is  to  5.     How  many  means  are  inserted  ? 

18.  Find  three  numbers  of  an  arithmetical  series  whose  sum 

shall  be  21,  and  the  sum  of  the  first  and  second  shall 
be  f  of  the  sum  of  the  second  and  third. 

19.  Find  three  numbers  whose  common  difference  is  1,  such 

that  the  product  of  the  second  and  third  exceeds  that 
of  the  first  and  second  by  h- 

20.  How  many  terms  of  the  series  1,  4,  7 must  be  taken, 

in  order  that  the  sum  of  the  first  half  may  bear  to 
the  sum  of  the  second  half  the  ratio  10  :  31  ? 

21.  A  travels  uniformly  20  miles  a  day  ;    B  starts  three 

days  later,  and  travels  8  miles  the  first  day,  12  the 
second,  and  so  on,  in  arithmetical  progression.  In 
how  many  days  will  B  overtake  A  ? 


SERIES.  307 

22.  A  number  consists  of  three  digits  wliicli  are  in  arith- 

metical progression  ;  and  this  number  divided  by  the 
sum  of  its  digits  is  equal  to  26  ;  but  if  198  be  added 
to  it,  the  digits  in  the  units'  and  hundreds'  places  will 
be  interchanged.     Required  the  number. 

23.  The  sum  of  the  squares  of  the  extremes  of  four  numbers 

in  arithmetical  progression  is  200,  and  the  sum  of  the 
squares  of  the  means  is  136.    What  are  the  numbers? 

24.  Show  that  if  any  even  number  of  terms  of  the  series  1, 

3,  5 be  taken,  the  sum  of  the  first  half  is  to  the 

sum  of  the  second  half  in  the  ratio  1:3. 

25.  A  and  B  set  out  at  the  same  time  to  meet  each  other 

from  two  places  343  miles  apart.  Their  daily  jour- 
neys are  in  arithmetical  progression,  A's  increase 
being  2  miles  each  day,  and  B's  decrease  being  5 
miles  each  day.  On  the  day  at  the  end  of  which 
they  met,  each  travelled  exactly  20  miles.  Find  the 
duration  of  the  journey. 

26.  Suppose  that  a  body  falls  through  a  space  of  I63V  feet  in 

the  first  second  of  its  fall,  and  in  each  succeeding  sec- 
ond 32^  more  than  in  the  next  preceding  one.  How 
far  will  a  body  fall  in  20  seconds  ? 

27.  The  sum  of  five  numbers  in  arithmetical  progression  is 

45,  and  the  product  of  the  first  and  fifth  is  |  of  the 
the  product  of  the  second  and  fourth.  Find  the 
numbers. 

28.  If  a  full  car  descending  an  incline  draw  up  an  empty 

one  at  the  rate  of  li  feet  the  first  second,  4^  feet  the 
next  second,  7i  feet  the  third,  and  so  on,  how  long 
will  it  take  to  descend  an  incline  150  feet  in  length  ? 
"What  part  of  the  distance  will  the  car  have  descended 
in  the  first  half  of  the  time  ? 


308  ALGEBRA. 


G-EOMETRICAL   SERIES. 

384.  A  series  is  called  a  G-eometrical  Series  or  a  G-eomet- 
rical  Progression  when  each  succeeding  term  is  obtained  by 
multiplying  the  preceding  term  by  a  constant  Tnultiplier. 

385.  The  general  representative  of  such  a  series  will  be 

a,  ar,  ar^,  ar^,  ar^ , 

in  which  a  is  the  first  term  and  r  the  constant  multiplier  or 
ratio. 

386.  Since  the  exponent  of  r  increases  by  1  for  every 
term,  the  exponent  will  always  be  1  less  than  the  number 
of  the  term  ;  so  that  the 

nth  term  =  ar""'^. 

387.  If  the  nth  term  be  denoted  by  I,  this  equation  be- 
comes  l  =  ar--\  (1) 

388.  The  geometrical  mean  between  two  numbers  is  the 
number  which  stands  between  them,  and  makes  with  them 
a  geometrical  series. 

389.  If  a  and  h  denote  two  numbers,  and  O  their  geo- 
metrical mean,  then,  by  definition  of  a  geometrical  series, 

a       Q' 
.-.  G  =  V^.  (2) 

390.  Sometimes  it  is  required  to  insert  several  geometri- 
cal means  between  two  numbers. 


SEEIES. 


309 


l^  m  =  the  number  of  means,  then  m-}~2  =  n,  the  whole 
number  of  terms ;  and  if  ??i  +  2  be  substituted  for  n  in  the 
equation  ^  ^  ^^,n-i 

the  result  is  1  =  ar*""^^, 

.  ^+1  ^  1  (3) 

a 
Thus,  if  it  be  required  to  insert  three  geometrical  means  between 
3  and  48,  the  value  of  r  ia  found  to  be 

3 

.•.r  =  2, 
and  the  series  will  be  3,  6,  12,  24,  48. 

391.  If  I  denote  the  last  term,  a  the  first  term,  n  the 
number  of  terms,  r  the  common  ratio,  and  s  the  sum  of  the 
n  terms,  then 

s  =  a  +  ar  +  ar^  +  «^  + -{-ar"*'^. 

Multiply  by  r,  rs  =  ar -j- a?^ -\- ai^ -{- +  a?'""-^  +  ar". 

Therefore,   by  subtracting   the   first  equation  from  the 

second,  _ 

'  rs  —  s  =  af  —  a, 

or  (r  —  1)  s  =  a  (r"  —  1), 

.  ,_a(r--l)  (^) 


r-1 

392.    When  r  is  <  1,  this  formula  will  be  more  convenient 
if  written 


393.    Since  .  l^ar""- 

rl  =  ar^. 


and  (4)  may  be  written 


rl 


a 


In  working  out  the  following  results,  the  student  will  make  use 
of  the  two  equation?,  l-=ar^-'^  and  s  =  ^^^   ~     . 


310 


ALGEBRA. 


No. 

1 
2 
3 
4 

5 
6 

7 


Given. 


a  r  n 

a  r  8 

a  n  s 

r  n  s 


Required. 


Results. 


._a  +  (r-l)s 
r 

I  (s  -  Z)"-i  -a{s-  af-i  =  0. 


1  = 


(r  —  1)  sr"-l 


yn- 

-1 

a(r»- 

1) 

r-] 
r-1' 

-"-^a« 

I 

a  r  I 
a  n  I 
r  n  I 


9 

10 
11 

12 

13 
14 
15 
16 

17 
18 
19 
20 


r  n 


r  n  s 
r  I  s 
n  I  s 


r^-l 
a  =  rZ  —  (r  —  1)  s. 

o  (s  -  a)"-i  ~l{s-  O**-^  =  0. 


a  n  I 
a  n  s 
a  I  s 
n  I  8 


'  n. 


s  s  —  a      ^ 

r« r  + =  0. 

a  a 


-I 


rn_. 


0. 


a  r  s 
a  I  8 
r  I  s 


n:-'°g;-'°«%l. 
log  r 

log  [g  +  (r  —  1)  s]  —  log  a 

~  logr 

log  I  —  los  a 


+  1. 


log  {s-a)-  log  (s  -  I) 
^^_  log? -log  [^^-(^-1)^]  +  1. 


logr 


SERIES. 


311 


Exercise  CXVII. 

L    Find  the  seventh  term  of  2,  6,  18 

sixth  term  of  3,  6,  12 

ninth  term  of  6,  3,  1 J 

eighth  term  of  1,  —  2,  4 

twelfth  term  of  x^,  x^,  x? 

fifth  term  of  4  a,  —^mo?,  ^m^a? 

2.  Find  the  geometrical  mean  between  IQx^y  and  30 0:3/^2. 

3.  Find  the  ratio  when  the  first  and  third  terms  are  5  and 

80  respectively. 

4.  Insert  two  geometrical  means  between  8  and  125 ;    and 

three  between  14  and  224. 

5.  If  a  =  2  and  r  =  3,  which  term  will  be  equal  to  162  ? 

6.  The  fifth  term  of  a  geometrical  series  is  48,  and  the  ratio 

2.     Find  the  first  and  seventh  terms. 

7.  Find  the  sum  of 

3  +  6  -}-  12  + to  eight  terms. 

1  —  3  +  9  — to  seven  terms. 

8  +  4  -f-  2  + to  ten  terms.  • 

.1  +  .5  +  2.5+ to  seven  terms. 

m r  +  7^~ to  live  terms. 

4       16 

8.  The  population  of  a  city  increases  in  four  years  from 

10,000  to  14,641.     What  is  the  rate  of  increase  ? 

9.  The  sum  of  four  numbers  in  geometrical  progression  is 

200,  and  the  first  term  is  5.     Find  the  ratio. 

10.    Find  the  sum  of  eight  terms  of  a  series  whose  last  term 
is  1,  and  fifth  term  i. 


312  ALGEBRA. 

11.  In  an  odd  number  of  terms,  show  that  the  product  of 

the  first  and  last  will  be  equal  to  the  square  of  the 
middle  term. 

12.  The  product  of  four  terms  of  a  geometrical  series  is  4, 

and  the  fourth  term  is  4.     Determine  the  series. 

13.  If  from  a  line  one-third  be  cut  off,  then  one-third  of  the 

remainder,  and  so  on,  what  fraction  of  the  whole  will 
remain  when  this  has  been  done  five  times  ? 

14.  Of  three  numbers  in  geometrical  progression,  the  sum 

of  the  first  and  second  exceeds  the  third  by  3,  and  the 
sum  of  the  first  and  third  exceeds  the  second  by  21. 
What  are  the  numbers  ? 

15.  Find  two  numbers  whose  sum  is  31  and  geometrical 

mean  H? 

16.  A  glass  of  wine  is  taken  from  a  decanter  that  holds  ten 

glasses,  and  a  glass  of  water  poured  in.  After  this 
is  done  five  times,  what  part  of  the  contents  is  wine? 

17.  There  are  four  numbers  such  that  the  sum  of  the  first 

and  the  last  is  11,  and  the  sum  of  the  others  is  10. 
The  first  three  of  these  four  numbers  are  in  arithmeti- 
cal progression,  and  the  last  three  are  in  geometrical 
progression.     Find  the  numbers. 

18.  Find  three  numbers  in  geometrical  progression  whose 

sum  is  13  and  the  sum  of  their  squares  91. 

19.  The  difference  between  two  numbers  is  48,  and  the  arith- 

metical mean  exceeds  the  geometrical  by  18.  Find 
the  numbers. 

20.  There  are  four  numbers  in  geometrical  progression,  the 

second  of  which  is  less  than  the  fourth  by  24,  and  the 
sum  of  the  extremes  is  to  the  sum  of  the  means  as  7 
to  3.     Find  the  numbers. 


SERIES. 


313 


21.  A  number  consists  of  three  digits  in  geometrical  pro- 
gression. The  sum  of  the  digits  is  13  ;  and  if  792  be 
added  to  the  number,  the  digits  in  the  units'  and 
hundreds'  places  will  be  interchanged.  Find  the 
number. 

394.  "When  r<  1,  a  geometrical  series  has  its  terms  con- 
tinually decreasing ;  and  by  increasing  n,  the  value  of  the 
nth  term,  a?-'*"-^  may  be  made  as  small  as  we  please,  though 
not  absolutely  zero. 

395.  The  formula  for  the  sum  of  n  terms, 

g  (1  -  r**) 
1-r 

may  be  written 


1  —  ?•      1  —  r 

By  increasing  n  indefinitely,  the  value  of becomes 

1  —  r 
indefinitely  small,  so  that  the  sum  of  7i  terms  approaches 

indefinitely  to      ^      as  its  limit. 
1  —  r 

Ex.  Find  the  limit  of  1  —  ^  +  i  —  i 


Here  a  =  l,  and  r  =  —  ^, 

1         ^     1    ,^2 

l-r      l-(-^)      1  +  i      3" 


Ans. 


22.    Find  the  limits  of  the  sums  of  the  following  infinite 
series : 

4  +  2  +  1+ 2-1I.  +  I- 

i  +  i  +  i+ .1  +  . 01 +  .001+ 

i-A  +  A- 868686 

l-f  +  ^- .54444 

t  +  iV  +  TV+ 83636 


314  ALGEBRA. 


*Haiimonical  Series. 

396.  A  series  is  called  a  Harmonical  Series,  or  a  Harmon- 
ical  Progression,  when  the  reciprocals  of  its  terras  form  an 
arithmetical  series. 

Hence,  the  general  representative  of  such  a  series  will  be 

1     _1_  1         ^  1 

a     a  +  c?'    a -{-2d  a-\-{n  —  l)d 

397.  Questions  relating  to  harmonical  series  should  be 
solved  by  writing  the  reciprocals  of  its  terms  so  as  to  form 
an  arithmetical  series. 

398.  If  a  and  h  denote  two  numbers,  and  ^  their  har- 
monical mean,  then,  by  the  definition  of  a  harmonical  series, 

2__i^i__l 

H     a      b      II 


.-. // 


Hah         ah 

2  ah 


a-^h 


399.  Sometimes  it  is  required  to  insert  several  harmoni- 
cal means  between  two  numbers. 

Ex.  Let  it  be  required  to  insert  three  harmonical  means 
between  3  and  18. 

Find  the  three  arithmetical  means  between  \  and  -^-g. 

These  are  found  to  be  if,  ^f,  y\;  therefore,  the  harmonical  means 
areiifl,^/;  or  3if,  o^/s. 

*  A  harmonical  series  is  so  called  because  musical  strings  of  uniform 
thickness  and  tension  produce  harmony  when  their  lengths  are  represented 
by  the  reciprocals  of  the  natural  series  of  numbers ;  that  is,  by  the  series, 
1,  i  h  h  h  ^t«- 


SERIES.  '  315 


Exercise  CXVIII. 

1.  Insert  four  harmonical  means  between  2  and  12. 

2.  Find  two  numbers  whose  difference  is  8  and  the  har- 

monical  mean  between  them  14. 

3.  Find  the  seventh  term  of  the  harmonical  series  3,  3f , 

4 

4.  Continue  to  two  terms  each  way  the  harmonical  series 

two  consecutive  terms  of  which  are  15,  16. 

5.  The  first  two  terms  of  a  harmonical  series  are  5  and  6. 

Which  term  will  equal  30? 

6.  The  fifth  and  ninth  terms  of  a  harmonical  series  are  8 

and  12.     Find  the  first  four  terms. 

7.  The  difi'erence  between  the  arithmetical  and  harmonical 

means  between  two  numbers  is  1|-,  and  one  of  the 
numbers  is  four  times  the  other.    Find  the  numbers. 

8.  Find   the    arithmetical,    geometrical,    and    harmonical 

means  between  two  numbers  a  and  h  ;  and  show  that 
the  geometrical  mean  is  a  mean  proportional  between 
the  arithmetical  and  harmonical  means.  Also,  ar- 
range these  means  in  order  of  magnitude. 

9.  The  arithmetical  mean  between  two  numbers  exceeds 

the  geometrical  by  13,  and  the  geometrical  exceeds 
the  harmonical  by  12.     What  are  the  numbers  ? 

10.  The  sum  of  three  terms  of  a  harmonical  series  is  11,  and 

the  sum  of  their  squares  is  49.     Find  the  numbers. 

11.  When  a,  h,  c  are  in  harmonical  progression,  show  that 

a:  c  :  :  a  —  h  -.h  —  c. 


CHAPTER  XXIL 

*  Choice.     Binomial  Theorem. 

400.  If  three  paths,  A,  B,  and  C,  lead  to  the  top  of  a 
mountain,  there  is  obviously  a  choice  of  three  different  ways 
of  ascending  the  mountain  ;  and  when  the  top  of  the  moun- 
tain is  reached,  there  is  again  a  choice  of  three  different 
ways  of  descending. 

How  many  different  ways  are  there  of  doing  both  ? 

If  a  traveller  ascend  by  A,  he  may  descend  hj  A,  B,  or 
C.  This  makes  three  ways  of  doing  both.  If  he  ascend  by 
B,  he  may  descend  by  A,  B,  or  C;  and  again,  if  he  ascend 
by  (7,  he  may  descend  by  ^,  ^,  or  C. 

Therefore,  there  are  3x3  =  9  ways  in  all  of  doing  both. 

These  ways  may  be  indicated  as  follows : 


1.    A  and  A. 

4. 

B  and  A. 

7. 

CandX 

2.    ^and^. 

5. 

B  and  B. 

8. 

Cand^. 

8.    A  and  C. 

6. 

B  and  a 

9. 

(7  and  C. 

401.  Suppose  the  traveller  does  not  wish  to  ascend  and 
descend  by  the  same  path,  then  what  choice  has  he  ? 

He  has  a  choice  of  three  different  ways  in  ascending. 

But  he  has  a  choice  of  only  two  ways  in  descending. 

If  each  of  the  three  ways  of  ascending  be  joined  to  the 
two  eligible  ways  of  descending,  the  result  is  3x2=6 
ways  of  doing  both, 

*  This  chapter  is  based  upon  Whitworth's  Choice  and  Chance,  and  manj 
of  the  examples  have  been  taken  from  that  elegant  work. 


CHOICE. 

317 

These  ways  are : 

1.    A  and  B. 

3.    Bs^ndA. 

5.    Ca^ndA. 

2.    A  and  C. 

4.    B  and  C. 

6.    Cand^. 

402.  If  a  box  contain  five  capital  letters,  A,  JB,  C,  D,  E, 
and  three  small  letters,  x,  y,  z,  in  how  many  different  ways 
may  a  capital  letter  and  a  small  letter  be  selected  ? 

A  capital  letter  may  be  selected  in  5  ways. 

With  each  capital  letter  selected,  a  small  letter  can  be 
joined  in  3  ways.  So  that  the  number  of  different  ways  in 
which  the  selection    can  be  made   is  3  X  5  =  15.      These 


ways  are  : 

Ax 

Bx 

Cx 

Dx 

Ex 

Ay 

Az 

By 
Bz 

Oy 

Cz 

Dy 
Dz 

Ey 

Ez 

403.    Hence  the  fundamental  principle  of  choice  : 

I.    If  one  thing  can  he  done  in  a  diffei'ent  ways,  and  {when 

it  has  been  done  in  any  one  of  these  ways)  another  thing  can 

he  done  in  b  different  ways,  then  h<yth  can  he  done  m  a  X  b  dif- 

ferent  ways. 

For,  corresponding  to  eacli  of  the  a  ways  of  doing  the  first  thing, 

there  are  h  ways  of  doing  the  second  thing.     Therefore,  altogether, 

there  are  a  X  5  ways  of  doing  both  things. 

(1)  On  a  shelf  are  7  English  and  5  French  books.     In  how 

many  ways  can  one  of  each  be  chosen  ? 
7x5  =  35.  Am. 

(2)  On  a  shelf  are  7  English,  5  French,  and  9  German 

books.     In  how  many  ways  can  two  books  be  chosen 

so  that  they  shall  be  in  different  languages  ? 

An  English  book  and  a  French  book  can  be  chosen  in  7  X  5 
— =  35  ways.  A  French  book  and  a  German  book  in  5  X  9  =  45 
WAys.    An  English  book  and  a  German  book  in  7x  9  =  63  ways. 

Hence,  there  is  a  choice  of  35  +  45  +  63  ==  143  ways.  Anz. 


318 


ALGEBRA. 


(3)  Out  of  8  different  pairs  of  gloves,  in  how  many  differ- 

ent ways  can  a  right-hand  and  a  left-hand  glove  be 

chosen  which  shall  not  form  a  pair  ? 

A  right-hand  glove  can  be  chosen  in  8  ways ;  and  when  it 
is  chosen  there  are  7  left-hand  gloves,  any  one  of  which  may 
be  put  with  it  without  making  a  pair.  Hence,  the  choice  is  in 
8  X  7  =  56  ways. 

(4)  Out  of  the  ten  digits,  0,  1,  2,  3,  4,  5,  6,  7,  8,  9,  how  many 

numbers  each  consisting  of  two  figures  can  be  formed  ? 

Since  0  has  no  value  in  the  left-hand  place,  the  left-hand  place 
can  be  filled  in  9  ways. 

The  right-hand  place  can  be  filled  in  10  ways,  since  repetitions 
of  the  digits  are  allowed  (as  22,  33,  etc.). 

Hence,  the  whole  number  is  9  X  10  ==  90. 

(5)  How  many  odd  numbers  consisting  of  two  figures  can 

be  formed  with  the  ten  digits  ? 

The  left-hand  place  can  be  filled  in  9  ways ;  the  right-hand 
place  in  only  5  ways,  since  it  must  be  either  1,  3,  5,  7,  or  9. 
Hence,  the  number  is*9  X  5=  45.   Ans. 

404.    By  a  simple  extension  of  Rule  I.  it  is  evident  that, 

II.  If  one  thing  can  he  done  in  a  ways,  and  then  a  second 
thing  can  he  done  in  b  ways,  then  a  third  in  c  ways,  then  a 
fourth  in  d  ways,  etc.,  the  numher  of  ways  of  doing  all  the 
things  will  5e  a  X  b  X  c  X  d,  etc. 

(1)    In  how  many  ways  can  four  Christmas  presents  be  given 
to  four  boys,  one  to  each  boy  ? 

The  first  present  may  be  given  to  any  one  of  the  boys ;  hence 
\    there  are  4  ways  of  disposing  of  it. 

The  second  present  may  be  given  to  any  one  of  the  other  three 
boys  ;  hence  there  are  3  ways  of  disposing  of  it. 

The  third  present  may  be  given  to  either  of  the  othoi  two 
boys ;  hence  there  are  2  ways  of  disposing  of  it. 


n 


CHOICE.  319 


The  fourth  present  must  be  given  to  the  last  boy ;  hence  there 
is  only  1  way  of  disposing  of  it. 

There  are,  then,  4x3x2x1  =  24  ways. 

(2)  In  how  many  ways  can  five  presents  be  given  to  two 

children  ? 

Each  present  may  be  disposed  of  in  2  ways ;  for  it  may  be  given 
to  either  child.  Hence,  the  whole  number  of  ways  of  giving  the 
presents  is  2x2x2x2x2  =  32. 

(3)  In  how  many  ways  can  five  presents  be  divided  between 

two  cliildren  ? 

This  question  differs  from  the  last  only  in  the  fact  that  a  divi- 
sion of  the  gifts  excludes  the  two  ways  in  which  either  child 
receives  all  the  gifts. 

Hence,  there  are  32  —  2  =  30  ways. 

C4)    In  how  many  ways  can  x  things  be  given  to  n  persons? 
n*.  Ans. 

(5)  In  how  many  ways  can  a  vowel  and  a  consonant  be 

chosen  out  of  the  alphabet  ? 

Since  there  are  in  the  alphabet  6  vowels  and  20  consonants,  a 
vowel  can  be  chosen  in  6  ways  and  a  consonant  in  20  ways,  and 
both  (Rule  I.)  in  6  x  20  =  120  ways. 

(6)  In  how  many  ways  can  a  two-lettered  word  be  made, 

containing  one  vowel  and  one  consonant  ? 

The  vowel  can  be  chosen  in  6  ways  and  the  consonant  in  20 
ways ;  and  then  each  combination  of  a  vowel  and  a  consonant 
can  be  written  in  2  ways ;  as  ac,  ca. 

Hence,  the  whole  number  of  ways  i3  6  X  20  X  2  =  240. 

405.  The  last  two  examples  show  the  difference  between 
a  selection  or  combination  of  different  things,  and  an  arrange- 
ment or  permutation  of  the  same  things. 


320 


ALGEBRA. 


Thus,  ac  form  a  combination  of  a  vowel  and  a  consonant,  and  ac 
and  ca  form  two  different  arrangements  of  this  combination. 

From  (5)  it  is  seen  that  120  different  combinations  can  be  made 
with  a  vowel  and  a  consonant ;  and  from  (6)  it  is  seen  that  240  diSer- 
eni  permutations  can  be  made  with  the  same. 

Again,  a,  &,  c  is  a  selection  of  three  letters  from  the  alphabet.  This 
selection  then  admits  of  6  different  arrangements  or  permutations,  aa 
follows : 

abc  bca  cab 

acb  bac  cba 

406.  A  selection  or  combination  of  any  number  of  ele- 
ments or  things,  means  a  group  of  that  number  of  elements 
or  things  put  together  without  regard  to  their  order  of 
sequence.  An  arrangement  or  permutation  of  any  number 
of  elements  or  things  means  a  group  of  that  number  of  ele- 
ments or  things  put  together  with  reference  to  their  order 
of  sequence. 

Arrangements  or  Permutations. 

407.  In  how  many  ways  can  the  letters  of  the  word 
Cambridge  be  arranged,  taken  all  at  a  time? 

There  are  nine  letters.  In  making  any  arrangement  aT[y  one  of 
the  letters  may  be  put  in  the  first  place.  Hence,  the  first  place  can 
be  filled  in  9  ways.  Then  the  second  place  can  be  filled  with  any 
one  of  the  remaining  eight  letters  ;  that  is,  in  8  ways. 

In  like  manner,  the  third  place  in  7  ways,  the  fourth  place  in  6 
ways,  and  so  on ;  and,  lastly,  the  ninth  place  in  1  way. 

If  the  nine  places  be  indicated  by  Eoman  numerals,  the  result 
(Bule  II.)  is  as  follows : 

I.  II.  III.  IV.  V.  VI.  VII.  VIII.  IX. 
9x8x7x6x5x4x  3   x    2   X   1=-  362,880  ways. 

408.  Hence,  it  will  be  seen  that, 

III.    The  number  cf  arrangements  or  permutations  ofn 
different  elements  or  things  talcen  all  at  a  time  is 
n(n~l)(n-  2)  (n  -  3) 3  X  2  X  1. 


CHOICE.  321 


For,  the  first  place  may  be  filled  in  n  ways,  then  the  second  place 
in  71  —  1  ways,  then  the  third  place  in  n  —  2  ways,  and  so  on  to  the 
last  place,  which  can  be  filled  in  only  1  way. 

Hence  (Rule  II.)  the  whole  number  of  arrangements  is  the  con- 
tinued product  of  all  these  numbers, 

n(7i-l)(n-2)(n-3) 3x2x1. 

409.  For  the  sake  of  brevity  this  product  is  written  in, 
and  is  read  factorial  n. 


410.    It  will  also  be  evident  that, 

IV.  The  number  of  arrangements  of  n  different  elefmenti, 
taken  i  at  a  time,  is 

n(n  —  l)(n  —  2) to  r  factors, 

thai  is,  n(n  —  1)  (n  —  2) [n  —  (r  —  1)], 

or  n(n—l)(n—2) (n  — r  +  1). 

For  the  first  place  can  be  filled  in  n  ways,  the  second  in  n  —  1 
ways,  the  third  place  in  n  —  2  ways,  and  the  rih  place  in  n  —  (r  —  1) 
ways. 

(1)  A  shelf  contains  4  English  books,  5  French  books,  and 

6  German  books ;  in  how  many  ways  can  these  books 
be  arranged? 

[15  =  1,307,674,368,000  ways. 

(2)  In  how  many  ways  can  these  books  be  arranged,  if  the 

books  of  each  language  be  kept  together  ? 

The  English  books  can  be  arranged  (Rule  III.)  in  [4  ways,  the 
French  books  in  [5  ways,  and  the  German  books  in  [6  ways. 
Also,  the  three  seU  can  be  arranged  in  |_3  different  orders.  Hence, 
the  number  of  ways  in  which  the  whole  can  be  done  is  (Rule  II.) 

[4xL£Xl6x[3-  12,441,600. 


322  ALGEBRA. 


(3)  Of  the  arrangements  possible  with  the  letters  of  the 

word  Cambridge : 

(i.)    How  many  will  begin  with  a  vowel  ? 
(ii.)   How  many  will  both  begin  and  end  with  a  vowel  ? 

In  filling  the  nine  places  of  any  arrangement  in  case  (i.),  the 
first  place  can  be  filled  in  only  3  ways,  the  other  places  in  [8 
ways. 

In  case  (ii.)  the  first  place  can  be  filled  in  3  ways,  the  last  place 
in  2  ways  (one  vowel  having  been  used),  and  the  remaining  seven 
places  in  [7  ways. 

Hence,  the  answer  to  (i.)  is  3  X  [8  =  120,960 ; 

to  (ii.)  is  3  X  2  X  [7  =  30,240. 

(4)  With  the  letters  of  the  word  Cambridge,  how  many 

arrangements  can  be  made  : 
(i.)   Each  beginning  with  the  word  Cam  ? 

(ii.)    Keeping  the  letters  Cam  always  together? 

For  case  (i.)  the  answer  is  evidently  [6  ;  since  our  only  choice 
lies  in  arranging  the  remaining  six  letters  of  the  word. 

Case  (ii.)  may  be  resolved  into  arranging  Cam  and  the  last  six 
letters,  regarded  as  seven  distinct  elements,  and  then  arranging 
the  letters  Cam. 

The  first  can  be  done  in  [7  ways,  and  the  second  in  |3  ways. 
Hence  (Rule  II.)  both  can  be  done  in  [7  X  [3  =  30,240  ways. 

(6)    In  how  many  ways  can  the  letters  of  the  word  Cam- 
bridge be  written : 
(i.)   Without  changing  the  place  of  any  vowel  ? 
(ii.)   Without  changing  the  order  of  any  vowel  ? 
(iii.)   Without  changing  the  relative  order  of  vowels 

and  consonants? 
In  (i.)  the  second,  sixth,  and  ninth  places  can  be  filled  each  in 
only  1  way  ;  the  other  places  in  [6  ways. 

Therefore,  the  whole  number  of  ways  is  [6,=  720. 
In  (ii.)  the  vowels  in  the  different  arrangements  are  always 
kept  in  the  order  a,  i,  e.    One  of  the  six  consonants  car  be  placed 
in  4  ways  :  before  a,  between  a  and  i,  between  i  and  e,  and  after  d 


CHOICE.  323 


Then  a  second  consonant  can  be  placed  in  5  ways,  a  third 
consonant  in  6  ways,  a  fourth  consonant  in  7  ways,  a  fifth  con- 
sonant in  8  ways,  and  the  last  consonant  in  9  ways.  Hence 
(Rule  II.)  the  whole  number  of  ways  is  4x5x6x7x8x9 
.    =60,480. 

In  (iii.)  the  vowels  can  be  arranged  in  [3  ways,  and  the  conso- 
nants in  [6  ways.  Hence  (Rule  II.)  the  number  of  ways  is 
[3  X  [6  =  4,320. 

(6)  In  how  many  ways  can  4  persons,  A,  B,  C,  D,  sit  at  a 

round  table  ? 

If  the  four  places  are  not  regarded  as  relative  to  each  other, 
then  the  whole  number  of  ways  is  [4  =  24.  But  if  the  four  places 
are  regarded  as  relative  to  each  other,  then  by  placing  one  as  A, 
in  one  position,  and  by  arranging  the  others  in  the  other  three 
positions,  the  whole  number  of  ways  is  [3  =  6. 

(7)  In  how  many  ways  can  6  persons  form  a  ring  ? 

Here  relative  position  is  required.  Hence,  the  whole  number 
.  of  ways  is  [5  =  120. 

(8)  How  many  three-lettered  words  can  be  made  from  the 

alphabet,  no  letter  being  repeated  in  the  same  word? 
26x25x24  =  15,600.  Am. 

(9)  How  many  four-lettered  words  ? 

26  X  25  X  24  X  23  =  358,800.  Ans. 

(10)  How  many  different  arrangements  can  be  made  of  the 

letters  in  the  word  eye  f    . 

Distinguish  the  e's  thus,  e^,  Cg,  a-nd  arrange  as  follows : 

«i  y  «2  h^-iV  y  H  «2 

62  3/  ei  ej  gj  2/  V  H^ 

These  six  arrangements  become  three  when  the  e's  are  not  dis- 
tinguished. That  is,  each  pair  of  arrangements  produced  by 
permuting  the  e's  is  reduced  to  a  single  arrangement. 

Hence,  tha.  number  of  different  arrangements  is  found  by  di- 
viding |_3  by  [2  ;  that  is,  by  dividing  the  number  of  arrangements 
possible  when  the  letters  are  all  different  by  the  number  of  ways 
in  which  the  two  c's  can  be  permuted. 


324  ALGEBRA. 

411.    In  how  many  different  orders  can  the  letters  a,  a,  a, 
X,  y  be  written  ? 

If  the  letters  were  all  different,  the  answer  (Rule  III.)  would  he  [5; 
but  the  three  o's  may  be  permuted  in  [3  =  6  ways. 

Hence,  the  [5  arrangements  may  be  divided  into  six  groups,  each 
group  constituting  but  a  single  arrangement  of  the  given  letters. 

[5 

Hence,  the  whole  number  of  given  orders  is  —  =  20. 

By  the  same  course  of  reasoning,  the  whole  number  of 
different  orders  of  a,  a,  a,  x,  x  is  found  to  be 

^    =10. 


[3^ 


412.    In  like  manner  for  any  other  numbers.     Hence, 
V.    The  number  of  arrangements  of  n  eleTnents,  of  which 
p  are  alike,  q  others  are  alike,  and  r  others  are  alike ,  is 

\n 


\e\i\l- 


(1)    In  how  many  ways  can  the  letters  of  the  word  Missis- 
sippi be  arranged  ? 

-J= — =34,560.  Ans. 
[4t4[2 


(2)  In  how  many  different  orders  can  a  row  of  4  white 

balls  and  3  black  balls  be  arranged  ? 

J=-  =  35.  Ans. 
[4(3 

(3)  In  how  many  ways  can  4  white  balls  and  3  black  balls 

be  placed  in  a  row,  if  the  balls  are  all  different  in 

size. 

17  =  5040.  Ans. 


CHOICE.  325 


413.  In  case  the  n  elements  to  be  arranged  are  all  differ- 
ent, but  repetitions  of  them  are  allowed,  then  in  making  any 
arrangement,  every  place  to  be  filled  can  be  filled  in  n  ways, 
since  we  may  always  repeat  an  element  already  used. 
Hence,  corresponding  to  Rules  III.  and  IV.,  the  following 
rules  apply  to  cases  in  which  repetitions  are  allowed  : 

VI.  The  number  of  arrangements  of  n  different  elements^ 
taken  all  at  a  time,  when  repetitions  are  allowed,  is  n". 

VII.  The  number  of  arrangements  of  n  different  elements, 
talcen  r  at  a  time,  when  repetitions  are  allowed,  is  n^. 

(1)  How  many  three-lettered  words  can  be  made  from  the 

alphabet,  when  repetitions  are  allowed  ? 
263  _  17  576_  ^^_ 

(2)  How  many  three-lettered  words  can  be  made  from  the 

6  vowels  when  repetitions  are  allowed  ? 

63  =  216.  An&. 

(3)  A  railway  signal  has  3  arms,  and  each  arm  may  take 

four  different  positions,  including  the  position  of  rest. 
How  many  signals  in  all  can  be  made  ? 

43  -  1  =  63.  Atis. 

(4;    In  the  common  system  of  notation,  how  many  numbers 
can  be  formed  consisting  of  not  more  than  5  digits  ? 

All  the  possible  numbers  may  be  regarded  as  consisting  of 
each  5  digits,  by  prefixing  zeros  to  the  numbers  consisting  of  less 
than  5  digits.     Thus,  247  may  be  written  00247. 

Hence,  every  possible  arrangement  of  5  digits  out  of  the  10 
digits  will  give  one  of  the  required  numbers  except  00000 ;  and 
the  answer  is  10*  — 1  =  99999  ;  that  is,  all  the  numbers  between 
0  and  100,000. 


326  ALGEBRA. 


(5)  With  the  digits  0,  1,  2,  3,  4,  5,  how  many  numbers  be- 
tween 1000  and  4000  can  be  formed? 

Here  we  have  to  fill,  in  each  possible  case,  four  places.  The 
first  place  can  be  filled  with  1,  2,  or  3,  that  is,  in  3  ways ;  the 
second,  third,  and  fourth  places  each  in  6  ways.  Therefore,  the 
answer  is  3x6^=  648. 

(6)  With  the  same  digits,  0,  1,  2,  3,  4,  5,  and  between  the 

same  limits,  1000  and  4000: 
(i.)   How  many  even  numbers  can  be  formed  ? 
(ii.)    How  many  odd  numbers  can  be  formed? 

(iii.)   How  many  numbers  divisible  by  5  ? 

Evidently  (i.)  and  (ii.)  are  like  Ex.  (5),  except  that  the  last  place 
can  be  filled  in  only  3  ways.  In  (i.)  the  last  place  must  be  filled 
by  0,  2,  or  4 ;  in  (ii.)  the  last  place  must  be  filled  by  1,  3,  or  5. 

Hence,  the  answer  in  each  of  these  cases  is  3x6x6x3  =  324. 

In  (iii.)  the  last  digit  must  be  either  0  or  5 ;  and  the  answer 
for  this  case  is  3x6x6x2  =  216. 

Combinations. 

414.  In  how  many  ways  can  3  vowels  be  selected  from 
the  5  vowels  a,  e,  i,  o,  u. 

The  number  of  ways  in  which  we  can  arrange  3  vowels  out  of  5  is 
(Rule  IV.)  5  X  4  X  3  =  60. 

These  60  arrangements  might  be  obtained  by  first  forming  all  the 
possible  selections  of  the  3  vowels  out  of  5,  and  then  arranging  the  3 
vowels  in  each  selection  in  as  many  ways  as  possible. 

The  3  vowels  of  each  selection  may  be  arranged  in  [3  =  6  ways. 

Hence  (Rule  II.), 

Number  of  selections  X  6  =  number  of  arrangements  =  60. 

Therefore,  number  of  selections  =  -^  =  10. 

415.  In  general, 

VIII.  Out  ofn  different  elements,  the  numher  of  selections 
of  r  elements  is  equal  to  the  numher  of  arrangements  of  n  ele* 
m£jits  divided  by  [r. 


n 


CHOICE. 


32^? 


For,  let  s  ^  number  of  ways  of  selecting  r  elements  out  of  n  ele- 
ments. Then  the  r.elements  thus  selected  may  be  arranged  (Rule  III.) 
in  [r  different  ways.  Therefore  (Rule  I.)  s  X  |_r  =  number  of  arrange- 
ments of  n  elements  taken  r  at  a  time. 

.      ^  number  of  arrangements 

[r 

(By  Rule  IV.)   The  numerator  of  this  fraction  is  equal  to 

n{n-l){n-2) [n-{r-l)]. 

_n{n  —  l){n  —  2) (n  —  r  +  1) 

■■■'  E 

If  both  terms  of  this  fraction  be  multiplied  by  |w  — r, 


the  result  is 


\r  [n_ 


If  n  —  r  =  p,  then  n  =p  +  r,  and  this  formula  may  be  written 

416.  The  value  of  this  fraction  is  not  altered  if  p  and  r 
be  interchanged.     Hence, 

IX.  Out  o/"  p  +  r  different  elements,  the  number  of  ways 
in  which  p  elements  can  he  selected  is  the  same  as  the  num- 
ber of  ways  in  which  r  elements  can  be  selected. 

Thus,  out  of  8  elements,  3  elements  can  be  selected  in  the  same  num- 
ber of  ways  as  5  elements ;  namely, 

18    _  8.7.6      _. 
^--^  =  56  ways. 

(1)    Out  of  20  consonants,  in  how  many  ways  can  18  be 
selected  ? 

The  18  can  be  selected  in  the  same  number  of  ways  as  2 ;  and 
the  number  of  ways  in  which  2  can  be  selected  (Rule  VIII.)  is 

20^  =  190.  ^„.. 
9. 


328  ALGEBRA. 


(2)  In  how  many  ways  can  the  same  choice  be  made  so  as 

always  to  include  the  letter  B  ? 

Taking  B  first  we  must  then  select  17  out  of  the  remaining  W 
consonants.     This  can  be  done  in 

19|J8  =  171  ways. 

(3)  In  how  many  ways  can  the  same  choice  be  made  so  as 

to  include  B  and  not  to  include  C? 

Taking  B  first,  we  have  then  to  choose  17  out  of  18,  0  being 
excluded.     This  can  be  done  in  18  ways. 

(4)  A  society  consists  of  50  members,  10  of  whom  are  phy- 

sicians.    In  how  many  ways  can  a  committee  of  6 
members  be  selected  so  as  to  include  1  physician. 
The  5  members  not  physicians  can  be  selected  in 
140 

and  the  1  physician  in  10  ways.    Hence,  the  6  can  be  selected  in 

140 
10  X       ,      different  ways. 
[5  [35 

f5)    In  how  many  ways  can  a  committee  of  6  members  be 
selected  so  as  to  include  at  least  one  physician? 
Six  members  can  be  selected  from  the  whole  society  in 
150 

Six  members  can  be  selected  from  the  whole  society,  so  as  to 
include  no  physician,  by  choosing  them  all  from  the  40  members 
who  are  not  physicians,  and  this  can  be  done  in 
[40 


[6   [34 


ways. 


|50  140  ,         ,  .     ,      . 

Hence,  -== ^=^-  =  number  of  ways  oi  selecting 

[6  [44     [6  [34  ^  ^ 

the  committee  so  as  to  include  at  least  one  physician. 


CHOICE.  329 


;6)  Out  of  20  Republicans  and  6  Democrats,  what  choice 
is  there  of  appointing  a  committee  consisting  of  3 
Republicans  and  2  Democrats  ? 

The  Republicans  can  be  selected  in -  =  1140  ways ; 

6x5  1x2x3 

and  the  Democrats  in  =15  ways.     Hence,  the  whole  cora- 

1X2 
mittee  can  be  appointed  in  1140  X  15  =  17,100  ways. 

C7)  From  20  Republicans  and  6  Democrats,  in  how  many 
ways  may  5  different  offices  be  filled,  three  of  which 
must  be  filled  by  Republicans,  and  the  other  two  by 
Democrats  ? 

The  first  three  offices  can  be  assigned  to  3  Republicans  in 
20  X  19  X  18  =  6840  ways  (Rule  II.) ;  and  the  other  two  offices 
can  be  assigned  to  2  Democrats  in  6  X  5  =  30  ways. 

There  is,  then,  a  choice  of  6840  X  30  -  205,200  different  ways. 


^8)  Out  of  20  consonants  and  6  vowels,  in  how  many  ways 
can  we  make  a  word  consisting  of  3  different  conso- 
nants and  2  difierent  vowels  ? 

20  X  19  X  18 
Three  consonants  can  be  selected  in — =  1140  ways 

6x5        -  1X2X3 

and  2  vowels  in  ■ =  15  ways.     Hence  (Rule  I.)  the  5  letters 

1  X  ^ 
can  be  selected  in  1140  X  15  =  17,100  ways. 

When  they  have  been  so  selected,  they  can  be  arranged  (Rule 
III.)  in  |5  =  120  different  orders.  Hence,  there  are  17,100  X 
120  =  2,052,000  different  ways  of  making  the  word. 

(9)  How  many  words  of  2  consonants  and  1  vowel  can  be 
formed  from  6  consonants  and  3  vowels,  the  vowel 
being  the  middle  letter  of  each  word  ? 

The  two  consonants  can  be  selected  in '15  ways ;  the  vowel  in 
3  ways.  Each  combination  of  the  2  consonants  and  1  vowel  can 
be  arranged  in  2x1x1  =  2  ways.  Hence,  the  number  of 
words  that  can  be  formed  is  15  X  3  X  2  =  90. 


330  ALGEBRA. 


(10)  How  many  words  of  8  consonants  and  3  vowels  can  be 

formed  from  the  alphabet,  if  one  of  the  vowels  is  to 

be  always  a  ? 

20  X  19  >^  18 
The  consonants  can  be  selected  in  - — — ~ —  =  1140  ways, 

^.1,  1-5x4      1^  1X2X3 

and  the  vowels  in  =  10  ways. 

1X2  ^ 

Then  the  six  letters  of  each  combination  can  be  arranged  in 
[6  =  720  ways.  Hence,  the  number  of  words  that  can  be  formed 
is  1140  X  10  X  720  =  8,208.000. 

417i  To  find  for  what  value  of  r  the  numher  of  selections 
of  n  things,  taken  i  at  a  time,  is  the  greatest. 

The  formula      ,  ^^  (n- l)(n-2) (n-r  +  1) 

lX2x3X r 

may  be  written     «  =  -X -X ■ — . 

1         2  3  r 

The  numerators  of  the  factors  on  the  right  side  of  this  equation 
begin  with  n,  and  form  a  descending  series  with  the  common  differ- 
ence 1;  and  the  denominators  begin  with  1,  and  form  an  ascending 
series  with  the  common  difference  1.  Therefore,  from  some  point  in 
the  series,  these  factors  become  less  than  1.  Hence,  the  maximum 
product  is  reached  when  that  product  includes  all  the  factors  greater 
than  1. 

When  n  is  an  odd  number,  the  numerator  and  the  denominator 

of  each  factor  will  be  alternately  both  odd  and  both  even ;  so  that 

the  factor  greater  than  1,  but  nearest  to  1,  will  be  the  factor  whose 

numerator  exceeds  the  denominator  by  2.     Hence,  in  this  case,  r  must 

have  such  a  value  that 

n  —  1 
71  —  r  +  l=r  +  2,      or      r  = 

When  n  is  an  even  number,  the  numerator  of  the  first  factor  will 
be  even  and  the  denominator  odd  ;  the  numerator  of  the  second  fac- 
tor will  be  odd  and  the  denominator  even ;  and  so  on,  alternately ; 
so  that  the  factor  greater  than  1,  but  nearest  to  1,  will  be  the  factor 
whose  numerator  exceeds  the  denominator  by  1.  Hence,  in  this  case, 
r  must  have  such  a  value  that 

n  — r  +  l  =  r  +  l,      or      ^^-• 


CHOICE.  331 


(1)  What  value  of  r  will  give  the  greatest  number  of  selec- 

tions out  of  7  things  ? 

Here  n  is  odd,  and  r  =  ^~    =     ~~    =  3. 
2  2 

1x2x3 

Ifr==4,then  ,^7x6x5x4^3^ 

1x2x3x4 

So  that,  when  the  number  of  things  is  odd,  there  will  be 
two  equal  numbers  of  selections ;  namely,  when  the  number 
of  things  taken  together  is  just  under  and  just  ovei'  one-half 
of  the  whole  number. 

(2)  What  value  of  r  will  give  the  greatest  number  of  selec- 

tions out  of  8  things  ? 

Here  n  is  even,  and  r  =  -  =  -  =  4. 
2     2 

.^^8X7X6X5^^    ^^,. 
1x2x3x4 

So  that,  when  the  number  of  things  is  even,  the  number 
of  selections  will  be  greatest  when  one-half  of  the  whoU 
are  taken  together. 

418.    It  may  be  shown  that, 

X.  The  numbef)'  of  ways  in  which  x-j-y  different  ele- 
ments can  be  divided  into  two  classes,  so  that  one  shall  con- 
tain X  and  the  other  j  elements,  is  equal  to  the  number  of 
ways  in  which  either  x  elements  or  j  elements  may  be  selected 
from  the  x  +  y  elements ;  or. 

For  each  division  of  a;  +  y  elements  into  two  classes,  one  consisting 
of  X  elements,  the  other  of  y  elements,  is  evidently  effected  by  making 
a  selection  of  x  elements  from  x  +  y  elements,  and  leaving  y  elements 
not  selected. 


332  ALGEBRA. 


In  like  manner  the  number  of  ways  in  which  x~[-y  -{-z 
different  elements  can  be  divided  into  3  classes,  containing 
X,  y,  and  z  elements  respectively,  is 

\x-\~y-\-z  T 

\=±=^===  ;  and  so  on. 

\x\]i\z 

(1)  In  how  many  ways  can  18  men  be  divided  into  2  classes 

of  6  and  12? 

•  [18 

(2)  In  how  many  ways  can  18  men  be  divided  into  2  groups 

of  9  each? 

According  to  the  rule,  the  answer  would  be 

The  two  groups,  considered  as  groups,  have  no  distinction ; 
therefore,  permuting  them  gives  no  new  arrangement,  and  the 
true  result  is  obtained  by  dividing  the  preceding  by  [2,  and  is 

[2[9[9 

If  any  condition  be  added  that  shall  make  the  two  groups  dif- 
ferent, as,  if  one  group  wear  red  badges  and  the  other  blue,  then 
the  answer  would  be  i  xg 

lH' 

419.  Whenever  the  groups  are  indifferent,  Rule  IX.  gives 
each  arrangement  repeated  as  many  times  as  the  groups 
can  be  permuted  one  with  another  ;  that  is,  |_2  when  there 
are  2  groups,  [3  when  there  are  3  groups,  and  so  on. 

He  nee,  the  result  found  by  the  rule  must  be  divided  by 
[2,  [3.  etc.,  in  order  to  obtain  the  true  result. 


CHOICE.  333 


(1)  In  how  many  ways  can  the  52  cards  in  a  pack  be  di- 

vided among  4  players,  each  to  have  13  ? 

Here  the  assignment  of  each  group  to  a  different  player  makes 
the  groups  different;  and  the  answer  is 
|52 
[13  [13  [13  [13' 

(2)  In  how  many  ways  can  the  52  cards  of  a  pack  be  di- 

vided into  4  piles? 

Here  the  groups  are  indifferent,  and  the  answer  is 
[52 
[4  123  123  [13  [13* 

(3)  A  boat's  crew  consists  of  8  men,  of  whom  2  can  row 

only  on  the  stroke  side  of  the  boat,  and  3  can  row 
only  on  the  bow  side.  In  how  many  ways  can  the 
crew  be  arranged  ? 

There  are  left  3  men  who  can  row  on  either  side ;  2  of  these 
must  row  on  the  stroke  side,  and  1  on  the  bow  side. 

The  number  of  ways  in  which  these  three  can  be  divided  is 

When  the  stroke  side  is  completed,  the  4  men  can  be  arranged 
in  [4  ways  ;  likewise,  the  4  men  of  the  bow  side  can  be  arranged 
in  [4  ways.  Hence  (Rule  II.)  the  arrangement  can  be  made  in 
3x[4[4=  1728  ways. 

(4)  In  how  many  ways  can  10  copies  of  Homer,  6  of  Vir- 

gil, and  4  of  Horace  be  given  to  20  boys,  so  that  each 
boy  may  receive  a  book  ? 

The  boys  have  to  form  themselves  into  a  group  of  10  for  Ho 
mer,  of  6  for  Virgil,  of  4  for  Horace.     This  can  be  done  m 

[20 


[10(6  [4 


different  ways. 


334  ALGEBRA. 


(5)    In  how  many  ways  can  3  copies  of  one  book,  2  of  an- 
other, and  1  of  a  third  be  given  to  a  class  of  12  boys, 
so  that  no  boy  shall  receive  more  than  1  book  ? 
In  every  possible  way  of  assigning  the  books,  6  boys  receive 
them.     These  6  may  be  selected  from  the  whole  12  (Rule  VIII.) 

ill  112 

rTTTT  ways. 
[6  [6       ^ 

When  thus  selected,  the  books  may  be  assigned  to  them  (Rule 
V.)  in  16 

Hence,  the  whole  number  of  ways  of  giving  the  books  is 
1JL2  [6 


.    16[6     [3[2[1 

(6)  In  how  many  ways  can  the  same  books  be  given  to  the 

12  boys,  so  that  no  boy  shall  receive  more  than  1 

copy  of  any  book  ? 

In  allotting  the  3  copies  of  the  first  book,  the  boys  are  to  be 
separated  into  two  groups  of  3  and  9 ;  and  the  group  of  3  will 
in  each  case  receive  the  books. 

112 
This  can  be  done  in  ,■ — —  ways. 
[3  [9       -^ 

Likewise  the  2  copies  of  the  second  book  may  be  given  in 

[12 

The  single  copy  of  the  third  book  can  be  given  in  12  ways. 
Hence  (Rule  II.)  the  books  may  be  given  in 

|12         |12 

-!=-  X    —    X  12  different  ways. 
[3  [9     [2  [10  ^ 

(7)  In  how  many  ways  can  2  letters  be  selected  from  a,  h, 

c,  d,  (?,/,  if  the  letters  may  be  repeated  in  making  the 

selection  ? 

"Without  repetitions,  2  letters  can  be  selected  from  6  in  15  ways. 
With  repetitions,  as  aa,  etc.,  6  selections  can  be  made.  Hence, 
there  are  15  +  6  =-  21  different  fselections. 


CHOICE.  335 


(8)    In  how  many  ways  can  selections  of  3  letters  be  made 
from  a,  h,  c,  d,  e,/,  if  letters  may  be  repeated? 
These  selections  will  be  of  3  kinds : 
(i.)  All  three  letters  different, 
(ii.)  Two  letters  alike,  the  third  different, 
(iii.)  All  three  letters  alike. 
(By  Rule  VIII.)  (i.)  gives  20  ways. 

(ii.)  gives  6  x  5  =  30 ;  for  we  can  choose  2  alike  in  6  ways, 
and  then  join  a  different  letter  to  each  pair  in  5  ways, 
(iii.)  gives  evidently  6  ways. 
Hence,  there  are  in  all  20  +  30  +  6  =  56  different  selections. 

420.    This  may  also  be  shown  as  follows : 

By  adding  the  6  letters,  a,  b,  c,  d,  e,  f,  to  each  selection  of 
the  kind  required,  they  will  become  selections  of  6  -f  3  =  9 
letters  out  of  6,  in  which  selections  each  letter  occurs  at 
least  once,  and  in  which,  therefore,  there  must  be  exactly  3 
repetitions.  These  repetitions  may  be  all  of  the  same  letter, 
or  divided  among  the  different  letters. 

Any  one  of  these  selections  may  be  made  by  writing  9 
places,  and  then  filling  them  in  alphabetical  order,  taking 
care  to  make  exactly  three  repetitions  in  passing  from  one 
end  of  the  row  to  the  other. 


Thus: 

1 

2         3         4         5         6         7         8 

a 
a 
a 

a 
b 
b 

a        a        b        c        d        e       f 
c        c        c        d        e         e       f 
c        c        d       d        e         ^       / 

By  striking  out  the  6  letters,  a,  b,  c,  d,  e,/,  each  once, 
there  is  left  in  the  first  row  aaa,  in  the  second  row  cce,  in  the 
third  row  cde ;  that  is,  three  of  the  required  selections. 

Now,  in  filling  the  9  places  for  each  row,  9  —  1  or  8  steps 
must  be  made,  of  which  exactly  3  are  repetitions,  and  each 
of  the  other  5  is  a  change  to  a  different  letter. 


336  ALGEBRA. 


The  3  repetitions  may  be  chosen  from  the  8  steps  (Rule 
VIII.)  in 

8x7x6      ^^ 

This,  then,  is  the  number  of  ways  in  which  3  letters  can 
be  selected  out  of  6  letters,  when  repetitions  are  allowed. 

In  other  words,  3  elements  can  be  selected  from  6  ele- 
ments, when  repetitions  are  allowed,  in  as  many  ways  as  3 
elements  can  be  selected  from  6  +  3  —  1,  or  8  elements 
without  repetitions. 

421.    Hence  the  general  rule  : 

XL  The  number  of  ways  of  selecting  r  elements  from  n 
different  elements,  when  repetitions  are  allowed,  is  the  same 
as  the  number  of  ways  of  selecting  r  elements  from  n  -f  r  —  1 
elements  without  repetitions. 

And  this  number  of  ways  is  (Rule  VIII.), 

\n  +  r-l      n(n-\-l)(n+2') (n  +  r-1) 

|?-|?2  —  1    ~  [r 

(1)  In  how  many  ways  can  4  elements  be  selected  from  n 

elements,  when  repetitions  are  allowed  ? 
n{n  +  l){n  +  2){n  +  3)    ^^^^ 

(2)  How  many  dominoes  are  there  in  a  set  numbered  from 

double  blank  to  double  nine  ? 

Each  domino  is  made  by  selecting  two  numbers  out  of  the  ten 
digits,  and  repetitions  are  included ;  that  is,  the  two  numbers  on 
a  domino  may  be  the  same. 

Hence,  the  number  of  dominoes  is  equal  to  the  number  of  i 
lections  of  2  from  10  +  2  —  1,  or  11,  without  repetitions. 

1X2 


CHOICE.  337 


(3)    In  how  many  ways  can  4  glasses  be  filled  with  6  kinds 
of  wine,  without  mixing  ? 

The  number  is  equal  to  the  number  of  ways  in  which  4  things 
can  be  selected  from  5  +  4  —  1,  or  8  things,  without  repetitions. 

5  X  6  X  7  X 


1x2x3x4 


70.  Ans. 


(4)    In  how  many  ways  can  6  rugs  be  selected  at  a  shop 
where  2  kinds  of  rugs  are  sold  ? 

The  number  is  equal  to  the  number  of  ways  of  selecting  6 
from  2  +  6  —  1  =  7,  without  repetitions. 

2x3x4x5x6x7^y 
1x2x3x4x5x6 
If  a  and  b  represent  the  two  kinds  of  rugs,  the  7  ways  are  aa 
follows  : 

a  a  a  a  a  a  a  a  ah  h  h 

a  a  a  a  ah  a.ahbhb 

a  a  a  a  b  b  a  b  b  h  b  b 

bl^bbbh 

422.    It  may  be  shown  that, 

XII.  The  number  of  ways  in  which  a  selection  {of  some,  or 
all)  can  be  Tnadefi^om  n  different  things  is  2"  —  1. 

For  each  thing  can  be  either  taken  or  left,  that  is,  can  be  disposed 
of  in  two  ways. 

There  are  n  things  ;  hence  (Rule  II.)  they  can  all  be  disposed  of  in 
2"  ways.  But,  among  these  ways  is  included  the  case  in  which  all 
are  rejected ;  and  this  case  is  inadmissable. 

Hence,  the  number  of  ways  of  making  a  selection  is  2"  —  1 . 

(1)  In  a  shop  window  20  different  articles  are  exposed  for 

sale.     What  choice  has  a  purchaser  ? 
220-1  =  1,048,575.  Ans. 

(2)  How  many  different  amounts  can  be  weighed  with  1  lb., 

2  lb.,  4  lb.,  8  lb.,  and  16  lb.  weights? 

2^-1  =  31.  Ans. 

(Let  the  student  write  out  the  31  weights.) 


338  ALGEBRA. 


423.    It  may  be  shown  that, 

XIII.    The  whole  number  of  ways  in  which  a  selection  can 

he  made  from  p-f-q  +  r things,  of  which  p  are  alike,  q  are. 

alilce,  r  are  alike,  etc.,  is  (p  + 1)  (^1+ 1)  (r  + 1) —  1. 

For  the  set  of  p  things  may  be  disposed  of  in  j?  +  1  ways,  since 
none  of  them  may  be  taken,  or  1,  2,  3 ,  orp,  may  be  taken. 

In  hke  manner,  the  q  things  may  be  disposed  of  in  5  +  1  ways ; 
the  r  things  in  r  +  1 ;  and  so  on. 

Hence  (Rule  II.)  all  the  things  may  be  disposed  of  in  {p  +  l)(g'  + 1) 
(r +  1) ways. 

But  the  case  in  which  all  the  things  are  rejected  is  inadmissable ; 
hence,  the  whole  number  of  ways  is  (p  +  l)(2'  +  l)(r  +  l) —  1. 

(1)  In  how  many  ways  can  2  boys  divide  betw^een  them  10 

oranges  all  alike,  15  apples  all  alike,  and  20  peaches 

all  alike? 

Here,  the  case  in  which  the  first  boy  takes  none,  and  the  case 
in  which  the  second  boy  takes  none,  must  be  rejected. 

Therefore,  the  answer  is  1  less  than  the  result,  according  to 
Rule  XIII.     11  X  16  X  21  -  2  =  3694.  Ans. 

(2)  If  there  be  m  kinds  of  things,  and  n  things  of  each  kind, 

in  how  many  ways  can  a  selection  be  made  ? 
In  this  case  p,  q,  r,  etc.,  are  all  equal,  and  each  is  equal  to  n. 
Hence,  the  result  is  {n  +  1)"*  —  1. 

QS)    If  there  be  m  kinds  of  things,  and  1  thing  of  the  first 
kind,  2  of  the  second,  3  of  the  third,  and  so  on,  in 
how  many  ways  can  a  selection  be  made  ? 
lm  +  1  —  1.  Ans. 

Exercise  CXIX. 

1.  How  many  different  permutations  can  be  made  of  the 

letters  in  the  word  Ecclesiastical,  taken  all  together  ? 

2.  Of  all  the  numbers  that  can  be  formed  with  four  of  the 

digits  5,  6,  7,  8,  9,  how  many  will  begin  with  56  ? 


CHOICE.  339 


3.  If  the  number  of  permutations  of  n  things,  taken  4  to- 

gether, "be  equal  to  12  times  the  permutations  of  n 
things,  taken  2  together,  find  n. 

4.  With  3  consonants  and  2  vowels,  how  many  words  of  3 

letters  can  be  formed,  beginning  and  ending  with  a 
consonant,  and  having  a  vowel  for  the  middle  letter  ? 

5.  Out  of  20  men,  in  how  many  different  ways  can  4  be 

chosen  to  be  on  guard  ?  In  how  many  of  these  would 
one  particular  man  be  taken,  and  from  how  many 
would  he  be  left  out. 

6.  Of  12  books  of  the  same  size,  a  shelf  will  hold  5.     How 

many  different  arrangements  on  the  shelf  may  be 
made? 

7.  Of  8  men  forming  a  boat's  crew,  one  is  selected  as  stroke. 

How  many  arrangements  of  the  rest  are  possible? 
When  the  4  who  ro^  on  each  side  are  decided  on, 
how  many  arrangements  are  still  possible  ? 

8.  How  many  signals  may  be  made  with  6  flags  of  differ- 

ent colors,  which  can  be  hoisted  either  singly,  or  any 
number  at  a  time  ? 

9.  How  many  signals  may  be  made  with  8  flags  of  differ- 

ent colors,  which  can  be  hoisted  either  singly,  or  any 
number  at  a  time  one  above  another  ? 

10.  How  many  different  signals  can  be  made  with  10  flags, 

of  which  3  are  white,  2  red,  and  the  rest  blue,  always 
hoisted  all  together  and  one  above  another? 

11.  How  many  signals  can  be  made  with  7  flags,  of  which 

2  are  red,  1  white,  3  blue,  and  1  yellow,  always  dis- 
played all  together  and  one  above  another  ? 

12.  In  how  many  different  ways  may  the  8  men  serving  a 

field-gun  be  arranged,  so  that  the  same  man  may  al- 
ways lay  the  gun  ? 


340  ALGEBKA. 


13.  Find  the  number  of  signals  which  can  be  made  with  4 

lights  of  different  colors  when  displayed  any  number 
at  a  time,  arranged  above  one  another,  side  by  side, 
or  diagonally. 

14.  From  10  soldiers  and  8  sailors,  how  many  different  par- 

ties of  3  soldiers  and  3  sailors  can  be  formed  ? 

15.  How  many  signals  can  be  made  with  3  blue  and  2 

white  flags,  which  can  be  displayed  either  singly,  or 
any  number  at  a  time  one  above  another  ? 

16.  In  how  many  ways  can  a  party  of  6  take  their  places 

at  a  round  table? 

17.  Out  of  12  Democrats  and  16  Republicans,  how  many 

different  committees  can  be  formed,  each  consisting 
of  3  Democrats  and  4  Republicans  ? 

18.  From  12  soldiers  and  8  sailors,  how  many  different  par- 

ties of  3  soldiers  and  2  sailors  can  be  formed  ? 

19.  Find  the  number  of  combinations  of  100  things,  97  to- 

gether ? 

20.  With  20  consonants  and  5  vowels,  how  many  different 

words  can  be  formed  consisting  of  3  different  conso- 
nants and  2  different  vowels,  any  arrangement  of 
letters  being  considered  a  word  ? 

21.  Of  30  things,  how  many  must  be  taken  together  in  or- 

der that  having  that  number  for  selection,  there  may 
be  the  greatest  possible  variety  of  choice  ? 

22i    There  are  ra  things  of  one  kind  and  n  of  another ;  how 

many  different  sets  can  be  made  containing  r  of  the 

first  and  s  of  the  second  ? 
23.    In  how  many  ways  may  10  persons  be  seated  at  a  round 

table,  so  that  in  no  two  of  the  arrangements  may 

every  one  have  the  same  neighbors? 


CHOICE.  341 


24.  The  number  of  combinations  of  n  things,  taken  r  to- 

gether, is  3  times  the  number  taken  r  —  1  together, 
and  half  the  number  taken  r  + 1  together.  Find  n 
and  r. 

25.  In  how  many  ways  may  12  things  be  divided  into  3 

sets  of  4  ? 

26.  How  many  words  of  6  letters  may  be  formed  of  3  vow- 

els and  3  consonants,  the  vowels  always  having  the 
even  places  ? 

27.  From  a  company  of  90  men,  20  are  detached  for  mount- 

ing guard  each  day.  How  long  will  it  be  before  the 
same  20  men  are  on  guard  together,  supposing  the 
men  to  be  changed  as  much  as  possible  ;  and  how 
many  times  will  each  man  have  been  on  guard  ? 

28.  Supposing  that  a  man  can  place  himself  in  3  distinct 

attitudes,  how  many  signals  can  be  made  by  4  men 
placed  side  by  side  ? 

29.  How  many  different  arrangements  may  be  made  of  11 

cricketers,  supposing  the  same  2  always  to  bowl  ? 

30.  Five  flags  of  different  colors  can  be  hoisted  either  singly, 

or  any  number  at  a  time  one  above  another.  How 
many  different  signals  can  be  made  with  them  ? 

31.  How  many  signals  can  be  made  with  5  lights  of  differ- 

ent colors,  which  can  be  displayed  either  singly,  or 
any  number  at  a  time  side  by  side,  or  one  above 
another  ? 

32.  The  number  of  permutations  of  n  things,  3  at  a  time,  is 

6  times  the  number  of  combinations,  4  at  a  time 
Find  n. 

33.  At  a  game  of  cards,  3  being  dealt  to  each  person,  any 

one  can  have  425  times  as  many  hands  as  there  are 
cards  in  the  pack.     How  many  cards  are  there  ? 


342  ALGEBRA. 


Binomial  Theorem. 
424.    By  performing  the  indicated  multiplication, 

(x -{- a)  (x -}- b)  =  x^  -'r  (a -}- b)  X -{-  ab. 
Multiply  each  side  hj  x-\-  c,  and  the  result  is 

(x-{-a)  {x-\-b)  (x-^c)  =  a^-{-  (a-\-b)  x^     +  oh^ 

+  co(?     -\-        (ac-{-bc)  x-\-abc 

==^  x^-\-  (a-\-b-\-c)  a^-j-  (ab-{-ac-\-bc)  x-\-abc 

From  this  result  certain  laws  are  to  be  observed : 

I.  The  number  of  terms  is  one  more  than  the  number  of  factors 
on  the  left  side. 

II.  The  exponent  of  x  in  the  first  term  is  the  same  as  the  number 
of  factors,  and  decreases  by  1  in  each  succeeding  term. 

III.  The  coefficient  of  x  in  the  first  term  is  unity  ; 
in  the  second  term,  the  sum  of  a,h,  c\ 

in  the  third  term,  the  sum  of  the  products  two  and  two,  ah,  ac,  he ; 
and  the  fourth  term  is  the  product  ahc. 

Do  the  same  laws  hold,  whatever  be  the  number  of  factors  ? 
Suppose  that  these  laws  hold  for  r  factors,  so  that 

{x  +  a){x  +  h) {x  +  m)  =  af  +p^3f-'^  +p^x''-^  +  p^af-^  + +  pr 

where  p,  stands  for  a  +  h  + +  m,  the  sum  of  the  second  terms  ; 

Pj  stands  for  ah  +  ac  + ,  the  sum  of  the  products,  two  and 

two; 

p,  stands  for  ahc  +  ahd  + ,  the  sum  of  the  products,  three  and 

three, 
p,  stands  for  ahcd ,  the  product  of  the  r  letters. 

Multiply  by  another  factor  {x  +  n),  and  the  product  of  the  r  +  1 
factors  is 

X^+^-\-     p^af       -f        Pz^'^         +        Pa^'^         + +       Pr^ 

+     nx*"       +      p^naf-"^     +      p^nsf-^    + +     pr-inx      +prn 

=X''+l+(Pj+?l)af+(p2+i>i^)a;*-^+(p3+P2^)^"^  + +{pr+Pr-V^)x+prn 

Here  the  laws  I.  and  II.  evidently, hold;  and  as  to  the  coefficients 
Pj  +n=a  +  h  + +  m  +  n,  the  sum  of  the  r  +  1  letters  ; 


BINOMIAL    THEOREM.  343 

jDg  +Pjn  =  {ab  +  ac  + )  +  {an  +  hn  + +  mn),  the  sum  of  the 

products,  two  and  two  ; 

products,  three  and  three. 

prU  =  ahc.  ..mn,  the  product  of  the  r  +  1  letters.     Hence, 

The  laws  hold  for  r  +  1  factors  if  they  hold  for  r  factors. 

But  they  have  been  shown  to  hold  for  three  factors,  therefore  they 

hold  for  four  factors,  and  therefore  for  Jive  factors  ;  and  so  on,  for  any 

number  of  factors.* 

425.    When  the  factors  in   the   preceding  proof  are  ail 

equal,  so  that  b,  c,  d, n,  are  each  equal  to  a,  the  left 

side  of  the  equation  becomes 

{x  +  a){x  +  a) taken  n  times  ;  that  is,  {x  +  a)**. 

On  the  right  side, 

p^  =  a  +  a  + =  a  taken  n  times  =  na  ; 

p^  =  aa  +  aa  + =  a^  taken  as  many  times  as  there  is  a  choice 

of  2  letters  from  n  letters, 

that  is,  »,  =  n{n-l)  ^g  .  ^^^ 

^*        1  X  2  ^ 

p^  =  aaa  +  aaa  + =  a^  taken  as  many  times  as  there  is  a  choice 

of  3  letters  from  n  letters. 

that  is,  «,  =  n{n-l){n-2)    ,  .  ^^^ 

^'  1x2x3  ^ 


Prn  =  ax  axa =  a  taken  n  times  as  a  factor  =  a". 

.-.  (x  +  a)«  =  x^  +  nax^-^  +  ^  ^^  ~  -^)  a^x"^-^ 
1x2 

,  n(n  —  l)(n~2)    -i  „  o 

1x2x3 

426.  The  expression  on  the  right  side  is  called  the  expan- 
sion of  (x  +  ay. 

If  a  and  x  be  interchanged,  the  expansion  will  proceed 
bv  ascending  powers  of  x,  as  follows  : 

La  +  a;)«  =  a»  +  na*'-'^x  +  '^\^~    )  a'^-^^  ^ +  nax^-^  +  a;". 

1x2 

*  A  proof  of  this  kind  is  called  mathematical  induction. 


344  ALGEBRA. 


Ifa  =  1,  then, 

(1  +  a;)«  =  1  +  nx  +  ^  ^^  ~]^^  x^  + +  7ia:«-i  +  a;«. 

i  X  -J 

If  a;  be  negative,  the  odd  powers  of  x  will  be  negative  and  the  tven 
powers  positive. 

(a  -  x)»  =  a«  -  ria«-i  x  +  ^  ^^^  ~  "^^  a«-2a;« 
1X2 

_n(n-l)(n-2)^,.3^     

1x2x3 


427.  It  will  be  observed  that  the  last  factor  in  the  de- 
nominator of  the  coefficient  is  1  less  than  the  number  of  the 
term,  and  is  the  same  as  the  exponent  of  the  second  letter ; 
also,  that  the  last  factor  of  the  numerator  of  the  coefficient 
is  found  by  subtracting  the  last  factor  in  the  denominator 
from  n-\-\,  and  that  the  exponent  of  the  first  letter  is  found 
by  subtracting  the  exponent  of  the  second  letter  from  n. 
So  that, 

The  rth  (or  general)  term  in  the  expansion  of  (a  +  xY  is 

n{n-\) (n-r  +  2)^„_,^,^,, 

1X2 (r-1) 

Thus,  the  third  term  of  {a  +  x)^  is 

20  X  19^18^^190^18^^ 
1X2 

428.  The  coefficient  of  the  rth  term  from  the  beginning 
is  equal  to  the  coefficient  of  the  rth  term  from  the  end. 

For,  the  coefficient  of  the  rth  term  from  the  beginning  is 

n(n-l) (n-r  +  2)  n{n-l) (n-r  +  2). 

1X2 r-1  '     ^^  |7^1  ' 

and  this  becomes,  when  both  terms  are  multiplied  by  |n  — r  +  1, 

{^ 

|r  —  1  \n  -r  -yl 


BINOMIAL    THEOREM.  345 

The  coefficient  of  the  rth  term  from  the  end,  which  is  the  {n—r+2)th 
term  from  the  beginning,  is 

n(n  — 1) r  . 

|n  — r+  1 

and  this  also  becomes,  when  both  terms  are  multiplied  by  [r  — 1, 

|r  —  1   \n  —r  +  1 

429.  It  will  be  evident  from  §  417,  that  in  the  expansion 
of  (a  +  xy,  the  middle  term  will  have  the  greatest  coeffi- 
cient when  n  is  even ;  and  when  n  is  odd  the  two  middle 
terms  will  have  equal  coefficients,  and  these  will  be  the 
greatest. 

Exercise  CXX. 
Expand  : 

1.  (l  +  2rr)^  3.    (2a:-3y/.  5.    (l-^Y 

2.  (37-3)8.  ^    (2-xf.  6.    (1--)'. 

7.  Find  the  fourth  term  of  (2:r  -  ^i/f^ 

8.  Find  the  seventh  term  of  f-  +  ^Y**- 

9.  Find  the  twelfth  term  of  (a^  -  ax)^. 
la  Find  the  eighth  term  of  (5:^7/  -  2x7/f, 
11.  Find  the  middle  term  of  f-  -f  ^' 


y     X. 

12,  Find  the  middle  term  of  f ^ 

X, 

13.  Find  the  two  middle  terms  of  f-  -  ^^^ 


346  ALGEBRA. 


14.  Find  the  o-th  term  of  (2  a  +  xf. 

15.  Find  the  rth  term  from  the  end  of  (2  a  +  a:)^ 

16.  Find  the  (r  +  ^)th  term  of  (a  +  xj'. 

17.  Find  the  middle  term  of  (a  -f  o:)^". 

18.  Expand  {2a-\-x)^,  and  find  the  sum  of  the  terms  if 


a=l,x  =  -2. 


When  the  Exponent  is  Fractional  or  Negative. 

430.    The  product  of 

l  +  ax  +  hx^  -f  cx^  + 

and  l  +  a'x  +  h'x^  +  dx^  + 

is  an  expression  in  ascending  powers  of  x,  as 

\-\-Ax^B3?'\-G2^^ , 

in  which  the  coefficients  A,  B,  C, are  functions  of  a,  b,  c, 

a\  b\  c\ ;  that  is,  are  made  up  of  these  letters  in  par- 
ticular ways. 

The  ways  in  which  a,  b,  c,  a\  b',  c, enter  into  these 

functions  will  evidently  be  the  same,  whatever  may  be  the 
values  assigned  to  the  letters. 

Likewise,  in  the  product  of 

l  +  mx^ ^^ — -^  x^  + 

1x2 

and  1  +  nx  +  ^  \    ~    ^  a^ -\- , 

1x2 

the  coefficients  of  a;,  xr^, will  be  functions  of  m  and  n,  the 

forms  of  which  will  be  the  same  for  all  values  of  m  and  n. 


I 


BINOMIAL    THEOREM.  347 


But  when  ra  and  n  are  positive  integers, 

{X+xT^X^mx^  "'f"~^-)  x'  + ;     §  425. 

i  X  -^ 


.-.  (1  +  :i-)"*  +  '*  ^  their  pro.duct ; 
and  (1  +  a:)"'  +  "  becomes,  by  expansion,  §  425. 

1  +  (?n  +  w)  :r  +  ^^ '- — ^ — ^ ^  3?  + 

These  forms,  then,  being  true  for  all  values  of  m  and  n, 
will  hold  when  m  and  n  2iXQ  fractional  or  negative. 

431.  If  the  expressions 

1  4-  mx  -\ ^^ ^  :r  + 

1x2 

and  \-\-nx-\ ^ ^a^-\- 

1X2         ^ 

be  represented  hyf(7n)  and  /(n),  their  product  will  be  rep- 
resented hyf{7n-\-n),  since  it  is  formed  with  m-f^  in  the 
same  way  as  they  are  formed  with  m  and  n. 

.•.f(m)xf(n)^f(m  +  n). 
In  like  manner, 

f(m)  Xf{n)  Xf(p)  ^f(m  +  n)  xf(p)  =f{m  +  n-\-p)- 
and  so  on  for  any  number  of  such  factors. 

432.  For  all  values  of  ?w,  n, 

/(m)  Xf(n) to  s  factors  =f(vi  +  n  + to  s  terms), 

and  if  m,  n, be  all  equal,  this  equation  becomes 

in  which  s  is  a  positive  integer. 


)48  ALGEBRA. 


Now,  if  ms  be  any  positive  integer  r,  so  that  m  becomes 

the  positive  fraction  -,  the  equation 

s 


becomes 


{/©}■-/«■ 

=  (1  +  xy,  since  r  is  integral. 

.•.(i+.)^-/g). 


e-) 


Hence,  the  /ottti  of  the  expansion  is  the  same  when  the 
exponent  is  a  positive  fraction  as  when  it  is  a  positive  integer. 

433.    Again,  since  the  equation 

f(7n)xf(n)=f(m  +  n) 
is  true  for  all  values  of  m  and  71,  it  is  true  when  n  =  —  7n. 
:.f(m)  x/(-  m)  =/(0),  which  equals  1.      §  255. 

that  is,  (1  +  ^)"'"  =/{-  ^0- 

Hence,  when  the  exponent  is  negative,  whether  integral 
or  fractional,  the  fo7^vi  of  the  expansion  is  the  same. 

Note.  In  the  expansion  of  (1  +  .t)'*,  if  n  is  a  positive  integer,  the 
numerator  of  the  last  factor  of  the  coefficient  of  the  (r  +  l)th  term, 
n  —  r  +  1,  will  be  equal  to  0  when  r  =  ?i  +  1 ;  this  term,  therefore,  and 
all  following  terms  (for  they  will  also  have  this  factor)  will  vanish. 
Hence,  the  series  will  end  with  the  rth  term.  But  if  n  is  fractional, 
or  negative,  no  value  of  r  will  make  n  —  r  +  1  =  0,  and  the  series  will 
be  infinite.  Hence,  the  sign  =  in  these  cases  will  mean,  ''is  equal  to 
the  limit  of  the  series" 


BINOMIAL    THEOREM.  349 


Expand  to  four  terms : 
(1)  (l  +  x)i. 

^l+lx-^a^  +  j\x^- 

1 


1  /„2     o„„x-i      -U,      2, 


(2)  -, 
'a'  —  'Zax 

Va^-2ax  ^  a   ^ 

-Ifll    ^    ,-H-^-l)/2:.Y     -H-i-l)(-|-2)/2a:Y) 
ail         2a  1x2         l,aj  1x2x3  \a  J  i 

ail         2a      8a2      IGa^  J 

A  root  may  often  be  extracted  bv  means  of  an  expansion. 

(3)  Extract  the  cube  root  of  344  to  six  decimal  places. 

.•.^34i  =  7(l+^ij)i, 

=7^i+ix^i^+i^i^(^i^)»+ y 

=  7  (1  +  .000971815  -  .000000944), 
=  7.006796. 

(4)  Extract  the  fifth  root  of  3128  to  six  decimal  places. 

3128  =  5^  +  3  =  5*/'l  +  i\ 

.••v^3l28  =  5^1+^V, 

1        5     55        1x2        \5')  r 

=  (1  +  .000192  -  .000000073728  + ), 

-  5.000959. 


350 


ALGEBRA. 


Exercise  CXXI. 


Expand  to  four  terms  : 
1.(1 -faOi  4.  (l-xy\  7.  .(2:r-33/)-i 

2.  (l+o;)!.  5.  (a' -0^)1  8.   </r^^^. 

3.  (a  +  x)^.  6.  (ar'i-xi/yl 


1 


10-  \L       o     5         11-  (l+^  +  ^)i 


V(4a2-3a:r/  \(l-3y/ 

12.  {l-x  +  x')l 

13.  Find  the  rth  term  of  (a  +  x)^. 

14.  Find  the  r?!A  term  of  (a  —  x)"^. 

15.  Find  VdS  to  five  decimal  places. 

16.  Find  V  1-jV  ^^  ^"^^  decimal  places. 

17.  Find  Vl29  to  six  decimal  places. 

18.  Expand  (1  —  2:r  -j-  Sx^)~i  to  four  terms. 

19.  Find  the  coefficient  of  x*  in  the  expansion  of  \    \^    (  • 

•       (1  +  32:^ 

20.  By  means  of  the  expansion  of  (1  +  xy  show  that  the 

limit  of  the  series 

1  +  1--^+  1X3 1x3x5  ^.....i3V2. 

2  2x2^  2x3x2»  2x3x4x2' 


